## State Competition Review Math Jam with Richard Rusczyk

Go back to the Math Jam ArchiveAs part of MATHCOUNTS Week, Art of Problem Solving’s Richard Rusczyk and David Patrick will discuss how to solve some of the toughest problems from the 2020 State MATHCOUNTS competition.

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#### Facilitator: Richard Rusczyk

rrusczyk
2020-05-14 15:14:08

It's time to get started!

It's time to get started!

rrusczyk
2020-05-14 15:14:17

I'm Richard Rusczyk. You might recognize me from the MATHCOUNTS Minis. My first MATHCOUNTS event was over 34 years ago. Yes, MATHCOUNTS really is that old.

I'm Richard Rusczyk. You might recognize me from the MATHCOUNTS Minis. My first MATHCOUNTS event was over 34 years ago. Yes, MATHCOUNTS really is that old.

rrusczyk
2020-05-14 15:14:19

(And, yes, that means I'm even older.)

(And, yes, that means I'm even older.)

rrusczyk
2020-05-14 15:14:34

The following year I made it to Nationals, representing Alabama.

The following year I made it to Nationals, representing Alabama.

rrusczyk
2020-05-14 15:14:42

Since then, I've returned to Nationals around 20 times. For the last 12-15 years, my colleague tonight (David Patrick,

Since then, I've returned to Nationals around 20 times. For the last 12-15 years, my colleague tonight (David Patrick,

**DPatrick**) and I spent the entire Registration Day at the registration desk visiting with students and old friends. If you have AoPS books in your home, you probably recognize DPatrick's name from some of the covers!
DPatrick
2020-05-14 15:14:45

Hi!

Hi!

RP3.1415
2020-05-14 15:14:57

yay!!

yay!!

Themiihacker
2020-05-14 15:14:57

hey

hey

menlo
2020-05-14 15:14:57

Hi!

Hi!

yekolo
2020-05-14 15:14:57

hi!

hi!

SandyBeach11
2020-05-14 15:14:57

Hi!

Hi!

genius_007
2020-05-14 15:14:57

Hello!

Hello!

bingo2016
2020-05-14 15:14:57

Hello!

Hello!

floatmeeting
2020-05-14 15:14:57

HI

HI

bobjoebilly
2020-05-14 15:14:57

Hi

Hi

MathWizard09
2020-05-14 15:14:57

hi!

hi!

math_piggy
2020-05-14 15:14:57

Hi!

Hi!

motao
2020-05-14 15:14:57

Hello

Hello

Abhinac
2020-05-14 15:14:57

hi

hi

LJCoder619
2020-05-14 15:14:57

Hi!

Hi!

rrusczyk
2020-05-14 15:15:12

We couldn't go to MATHCOUNTS this year. We were going to miss you all so much that we partnered with MATHCOUNTS to bring MATHCOUNTS to us.

We couldn't go to MATHCOUNTS this year. We were going to miss you all so much that we partnered with MATHCOUNTS to bring MATHCOUNTS to us.

PI-EATEN
2020-05-14 15:15:25

Yay

Yay

SparklyFlowers
2020-05-14 15:15:25

Yay! I'm so excited.

Yay! I'm so excited.

LANSH
2020-05-14 15:15:25

yay!

yay!

sdattilo2002
2020-05-14 15:15:25

YAY!!!!!

YAY!!!!!

superls
2020-05-14 15:15:25

yay!

yay!

guo35140
2020-05-14 15:15:25

3 14!!!!!

3 14!!!!!

Piano_Man123
2020-05-14 15:15:25

Yay!

Yay!

Layla2018
2020-05-14 15:15:25

YAY!!

YAY!!

Mathematician1010
2020-05-14 15:15:32

yay!

yay!

Kesav
2020-05-14 15:15:32

awesome

awesome

KLBBC
2020-05-14 15:15:32

Yeah

Yeah

jellybeanchocolate
2020-05-14 15:15:37

yay!

yay!

bedwinprusik578
2020-05-14 15:15:37

yay

yay

INDYMATHz
2020-05-14 15:15:37

nice

nice

LAC08512
2020-05-14 15:15:37

YAY!

YAY!

rrusczyk
2020-05-14 15:15:48

We have two special guests this afternoon from MATHCOUNTS. I've had the great fortune to work with MATHCOUNTS for even longer than this website has existed, and I've always been amazed at what they pull off each year.

We have two special guests this afternoon from MATHCOUNTS. I've had the great fortune to work with MATHCOUNTS for even longer than this website has existed, and I've always been amazed at what they pull off each year.

rrusczyk
2020-05-14 15:16:04

Many of you here today know about the Competition Series, but you might not have thought of how much work goes into making contests happen. Or the fact that it needs to happen around 600 times. Staffed almost entirely with volunteers. It still blows my mind that they've been able to make that work for nearly 40 years now.

Many of you here today know about the Competition Series, but you might not have thought of how much work goes into making contests happen. Or the fact that it needs to happen around 600 times. Staffed almost entirely with volunteers. It still blows my mind that they've been able to make that work for nearly 40 years now.

rrusczyk
2020-05-14 15:16:13

On top of that, they keep adding more, like the Club Program and the Math Video Challenge that many of you have been part of in the past few years.

On top of that, they keep adding more, like the Club Program and the Math Video Challenge that many of you have been part of in the past few years.

rrusczyk
2020-05-14 15:16:24

Representing MATHCOUNTS today, we have Executive Director Kristen Chandler (

Representing MATHCOUNTS today, we have Executive Director Kristen Chandler (

**LadyBlue32**) and Senior Manager of Education Kera Johnson (**KDEJ22**). If you took the test earlier this week, you saw Kristen and her indoor pool in the intro video before the test.
mathw0nder
2020-05-14 15:17:24

Hi Kristen and Kera!

Hi Kristen and Kera!

bestzack66
2020-05-14 15:17:24

Hello!

Hello!

floatmeeting
2020-05-14 15:17:24

hi!

hi!

Captus
2020-05-14 15:17:24

Hi!

Hi!

sixoneeight
2020-05-14 15:17:24

Hey!

Hey!

OlympusHero
2020-05-14 15:17:24

Wow, this is awesome! Hi Kristen and Kera!

Wow, this is awesome! Hi Kristen and Kera!

MathAcademyIsFun
2020-05-14 15:17:24

hi!

hi!

CSPAL
2020-05-14 15:17:24

Hello!!!

Hello!!!

MathAcademyIsFun
2020-05-14 15:17:24

hi

hi

zhuqingzhang
2020-05-14 15:17:24

hi

hi

kdej22
2020-05-14 15:17:30

hello!

hello!

kdej22
2020-05-14 15:18:10

glad to be here

glad to be here

rrusczyk
2020-05-14 15:18:30

Next let's talk about how the classroom works.

Next let's talk about how the classroom works.

LadyBlue32
2020-05-14 15:18:45

Hey everyone! We're thrilled to be joining you!! Thanks for making MATHCOUNTS Week so successful@

Hey everyone! We're thrilled to be joining you!! Thanks for making MATHCOUNTS Week so successful@

nnnF1s
2020-05-14 15:19:04

Thank you

Thank you

Hard
2020-05-14 15:19:04

Thank you! I was awesome!

Thank you! I was awesome!

tkfun
2020-05-14 15:19:04

hello! thanks!

hello! thanks!

Noam2007
2020-05-14 15:19:04

Thank you!

Thank you!

Cozzmo
2020-05-14 15:19:09

Yay! thx 4 being here

Yay! thx 4 being here

ObjectZ
2020-05-14 15:19:10

You're welcome!

You're welcome!

MathJams
2020-05-14 15:19:14

Thanks for doing this

Thanks for doing this

jabuticaba149
2020-05-14 15:19:22

YAY!!

YAY!!

Sports8531
2020-05-14 15:19:22

Thanks

Thanks

s0arbeacon
2020-05-14 15:19:22

No, Thank yOU!

No, Thank yOU!

fasterthanlight
2020-05-14 15:19:22

It's an honor to meet you!

It's an honor to meet you!

rrusczyk
2020-05-14 15:19:26

As you've probably figured out by now, the classroom is

As you've probably figured out by now, the classroom is

**moderated**, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
rrusczyk
2020-05-14 15:19:36

This helps keep the class organized and on track. This also means that only

This helps keep the class organized and on track. This also means that only

**well-written**comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
rrusczyk
2020-05-14 15:19:42

You can adjust the screen sizing in a variety of ways; give it a try! Also, you can change the font size by clicking the "A"s at the top of the room. Every few years I have to click the big A one more time, sigh.

You can adjust the screen sizing in a variety of ways; give it a try! Also, you can change the font size by clicking the "A"s at the top of the room. Every few years I have to click the big A one more time, sigh.

rrusczyk
2020-05-14 15:19:56

There are bunches and bunches of students here. As I said, only a fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it.

There are bunches and bunches of students here. As I said, only a fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it.

**I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here today**!
rrusczyk
2020-05-14 15:20:11

For those keeping score at home, we have over 1100 people in the room right now.

For those keeping score at home, we have over 1100 people in the room right now.

rrusczyk
2020-05-14 15:20:17

In our regular classes, there are (far, far) fewer students, and everyone gets plenty of "air time" if they participate.

In our regular classes, there are (far, far) fewer students, and everyone gets plenty of "air time" if they participate.

rrusczyk
2020-05-14 15:20:26

Also, we won't be going through the math quite as thoroughly as we do in our classes -- we can't teach all the prerequisite material for every problem as we go.

Also, we won't be going through the math quite as thoroughly as we do in our classes -- we can't teach all the prerequisite material for every problem as we go.

**Another difference between today and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a very, very large number of students today!**
rrusczyk
2020-05-14 15:20:42

We do have three more special people to introduce.

We do have three more special people to introduce.

rrusczyk
2020-05-14 15:20:45

Helping us here today, we have three assistants: Zach Stein-Perlman (

Helping us here today, we have three assistants: Zach Stein-Perlman (

**ZachSteinPerlman**), Agustin Marchionna (**tuvie**), and Bogdan Blaga (**nasrl**).
rrusczyk
2020-05-14 15:20:51

Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.

Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.

ZachSteinPerlman
2020-05-14 15:20:56

Hi!

Hi!

AOPSmathematics
2020-05-14 15:21:07

Hi!!!

Hi!!!

EvanZ
2020-05-14 15:21:07

hello!

hello!

floatmeeting
2020-05-14 15:21:07

Hi!

Hi!

mathcounting
2020-05-14 15:21:07

hi

hi

memi
2020-05-14 15:21:07

hi!!!!

hi!!!!

jupiter314
2020-05-14 15:21:07

hello

hello

Heavytoothpaste
2020-05-14 15:21:07

hi

hi

ElNadia
2020-05-14 15:21:07

HI

HI

s979903
2020-05-14 15:21:07

Hello!

Hello!

Lord_Vader
2020-05-14 15:21:07

hi

hi

GrantStar
2020-05-14 15:21:07

hello!

hello!

Chesssaga
2020-05-14 15:21:07

Hi!

Hi!

ObjectZ
2020-05-14 15:21:07

Howdy!

Howdy!

G1-100
2020-05-14 15:21:07

Hi

Hi

Ilikeminecraft
2020-05-14 15:21:07

hi

hi

falcon22
2020-05-14 15:21:07

hi!

hi!

Stunji_Ray
2020-05-14 15:21:07

hi

hi

OlympusHero
2020-05-14 15:21:15

Hi Zach!

Hi Zach!

PureSwag
2020-05-14 15:21:15

Hi Zach!!!

Hi Zach!!!

math_piggy
2020-05-14 15:21:15

Hi Zach!

Hi Zach!

Dumdum990
2020-05-14 15:21:15

Hello Zach!

Hello Zach!

aaravdodhia
2020-05-14 15:21:15

Hey Zach!

Hey Zach!

rrusczyk
2020-05-14 15:21:17

Agustin joined AoPS recently in 2019. He discovered his passion for maths when he first attended a math olympiad in 5th grade. Since then, he participated in many math olympiads, obtaining several prizes, such as a bronze medal in the IMO. After graduating from high school, he developed a new passion for programming contests. He is attending the University of Buenos Aires for a BS and MS in maths and computer science. He now helps in the organization of the Argentinian Math Olympiad, including tutoring and grading, giving back what he has been given during his time as olympian. In his spare time, Agustin loves to play soccer, as well as soccer related video games.

Agustin joined AoPS recently in 2019. He discovered his passion for maths when he first attended a math olympiad in 5th grade. Since then, he participated in many math olympiads, obtaining several prizes, such as a bronze medal in the IMO. After graduating from high school, he developed a new passion for programming contests. He is attending the University of Buenos Aires for a BS and MS in maths and computer science. He now helps in the organization of the Argentinian Math Olympiad, including tutoring and grading, giving back what he has been given during his time as olympian. In his spare time, Agustin loves to play soccer, as well as soccer related video games.

tuvie
2020-05-14 15:21:23

Hi everyone!

Hi everyone!

douglasubella
2020-05-14 15:21:39

Hi Agustin!!

Hi Agustin!!

Possible
2020-05-14 15:21:39

Hey Agustin!

Hey Agustin!

MathWizard09
2020-05-14 15:21:39

Hi Agustin!

Hi Agustin!

DaBobWhoLikeMath1234
2020-05-14 15:21:39

Hi Agustin!

Hi Agustin!

CosmoMonkhouse
2020-05-14 15:21:41

Hi Agustin!

Hi Agustin!

Speedstorm
2020-05-14 15:21:43

Hi Agustin.

Hi Agustin.

akmathisfun2020
2020-05-14 15:21:43

Hi Agustin!!!

Hi Agustin!!!

rrusczyk
2020-05-14 15:21:47

Bogdan is finishing his second year at Oxford University, studying Mathematics and Computer Science. He's been passionate about mathematics ever since his grandpa showed him the multiplication so that he can help with keeping scores at card games. He plans on pursuing a PhD, probably in pure math, but there are so many cool things to study! Outside of school, he really enjoys playing video games, basketball, and buying as many colorful socks as possible.

Bogdan is finishing his second year at Oxford University, studying Mathematics and Computer Science. He's been passionate about mathematics ever since his grandpa showed him the multiplication so that he can help with keeping scores at card games. He plans on pursuing a PhD, probably in pure math, but there are so many cool things to study! Outside of school, he really enjoys playing video games, basketball, and buying as many colorful socks as possible.

Mathworker2705
2020-05-14 15:22:12

Hi Bogdan!!!

Hi Bogdan!!!

KNM
2020-05-14 15:22:12

Hello Bogdan!

Hello Bogdan!

vanigarcha
2020-05-14 15:22:12

Hey Bogdan!

Hey Bogdan!

MathWizard38025
2020-05-14 15:22:12

Hey Bogdan!

Hey Bogdan!

Mandakini
2020-05-14 15:22:12

Hiiii

Hiiii

CHIPPER33
2020-05-14 15:22:12

Bogdon!!

Bogdon!!

Handiccraft
2020-05-14 15:22:12

hi bogdan!

hi bogdan!

Mathematician1010
2020-05-14 15:22:12

Hi Bogdan!

Hi Bogdan!

Elaine_Wang
2020-05-14 15:22:12

Hi!

Hi!

goodskate
2020-05-14 15:22:12

Hi Bogdan!

Hi Bogdan!

Piano_Man123
2020-05-14 15:22:12

HI BOgdan!!!!!

HI BOgdan!!!!!

areevv
2020-05-14 15:22:12

hi bogdan!!

hi bogdan!!

YuryK
2020-05-14 15:22:17

Hi Bogdan!

Hi Bogdan!

IceNeptune
2020-05-14 15:22:17

Hey Bogdan!!!

Hey Bogdan!!!

Facejo
2020-05-14 15:22:17

HELLO BOGDAN!!

HELLO BOGDAN!!

rrusczyk
2020-05-14 15:22:31

So, please go ahead and ask questions, but also please understand if we aren't able to answer them all! Our star assistants will get to as many as they can!

So, please go ahead and ask questions, but also please understand if we aren't able to answer them all! Our star assistants will get to as many as they can!

rrusczyk
2020-05-14 15:22:39

Today we're going to work through some of the hardest problems on this week's competition. I know I won't be able to keep up with you through the whole Math Jam, so

Today we're going to work through some of the hardest problems on this week's competition. I know I won't be able to keep up with you through the whole Math Jam, so

**DPatrick**is going to join me in working through the problems with all of you.
Handiccraft
2020-05-14 15:22:52

can we start?

can we start?

rrusczyk
2020-05-14 15:22:56

Good idea!

Good idea!

rrusczyk
2020-05-14 15:23:02

We'll start with the final six problems from the Sprint Round.

We'll start with the final six problems from the Sprint Round.

**No calculators allowed!**
rrusczyk
2020-05-14 15:23:05

**Sprint #25:**Equilateral triangle $ABC$ with side length $11$ is constructed on a triangular lattice grid with $12$ lattice points on each side. One of the lattice points inside the triangle is randomly selected and labeled point $P.$ If $a, b$ and $c$ are the distances from $P$ to sides $BC, AC,$ and $AB,$ respectively, as shown, what is the probability that a triangle exists with side lengths $a, b,$ and $c?$ Express your answer as a common fraction.
rrusczyk
2020-05-14 15:23:13

rrusczyk
2020-05-14 15:23:26

Note that you can adjust the bar between the top section and the main room.

Note that you can adjust the bar between the top section and the main room.

rrusczyk
2020-05-14 15:23:45

You can also double-click on images to pop them out into their own windows.

You can also double-click on images to pop them out into their own windows.

arirah9
2020-05-14 15:23:48

this was actually quite hard

this was actually quite hard

rrusczyk
2020-05-14 15:24:10

A lot of these problems were quite hard. My theory is that the test writers forgot that Luke and Andrew aren't in middle school anymore

A lot of these problems were quite hard. My theory is that the test writers forgot that Luke and Andrew aren't in middle school anymore

romani
2020-05-14 15:24:31

How do you solve it?

How do you solve it?

rrusczyk
2020-05-14 15:24:35

Great question. How do we usually tackle probability problems like this?

Great question. How do we usually tackle probability problems like this?

IceNeptune
2020-05-14 15:25:10

you need desired/total

you need desired/total

menlo
2020-05-14 15:25:10

success/all possibilites

success/all possibilites

falcon22
2020-05-14 15:25:21

Successful Outcomes/Total Possible Outcomes

Successful Outcomes/Total Possible Outcomes

sv739
2020-05-14 15:25:23

successful outcomes/ all possibilites

successful outcomes/ all possibilites

rrusczyk
2020-05-14 15:25:26

We need to compute $\dfrac{\text{number of points that work}}{\text{total number of points}}$.

We need to compute $\dfrac{\text{number of points that work}}{\text{total number of points}}$.

rrusczyk
2020-05-14 15:25:40

This should work because the points are equally likely to be chosen.

This should work because the points are equally likely to be chosen.

rrusczyk
2020-05-14 15:25:47

Let's start with the denominator. How many total interior lattice points are there to choose from?

Let's start with the denominator. How many total interior lattice points are there to choose from?

CT17
2020-05-14 15:26:27

There are $\frac{9\cdot 8}{2} = 45$ total points.

There are $\frac{9\cdot 8}{2} = 45$ total points.

CHIPPER33
2020-05-14 15:26:27

45

45

fasterthanlight
2020-05-14 15:26:27

45

45

walrus987
2020-05-14 15:26:27

45

45

KevinW12345
2020-05-14 15:26:27

45

45

MTHJJS
2020-05-14 15:26:27

1+2+3+4+5+6+7+8+9=45

1+2+3+4+5+6+7+8+9=45

WAWATHEGOAT
2020-05-14 15:26:27

9+8+7+6... = 45

9+8+7+6... = 45

Mrs. H math teacher
2020-05-14 15:26:27

1 + 2 + 3 + . . . + 9

1 + 2 + 3 + . . . + 9

rrusczyk
2020-05-14 15:26:52

I see a lot of you answered 78 here -- have to be careful!

I see a lot of you answered 78 here -- have to be careful!

rrusczyk
2020-05-14 15:27:04

We are choosing from the points

We are choosing from the points

**inside**the triangle!
rrusczyk
2020-05-14 15:27:10

There's $1$ point in the top row (nearest $A$), then $2$ points in the next row, and so on.

There's $1$ point in the top row (nearest $A$), then $2$ points in the next row, and so on.

rrusczyk
2020-05-14 15:27:14

skyleristhecoolest
2020-05-14 15:27:19

the sides don't count

the sides don't count

jacobwu
2020-05-14 15:27:19

the edge doesn't count

the edge doesn't count

rrusczyk
2020-05-14 15:27:25

Exactly!

Exactly!

rrusczyk
2020-05-14 15:27:29

Adding the number of lattice points inside $\triangle ABC$ in each horizontal row, there are \[1+2+3+4+5+6+7+8+9=45\]lattice points from which we can choose $P.$

Adding the number of lattice points inside $\triangle ABC$ in each horizontal row, there are \[1+2+3+4+5+6+7+8+9=45\]lattice points from which we can choose $P.$

rrusczyk
2020-05-14 15:27:34

Next, we count the number of points $P$ that work. Those for which the distances $a,b,c$ could possibly be the side lengths of a triangle.

Next, we count the number of points $P$ that work. Those for which the distances $a,b,c$ could possibly be the side lengths of a triangle.

rrusczyk
2020-05-14 15:27:41

This is where it gets tricky!

This is where it gets tricky!

rrusczyk
2020-05-14 15:27:46

How might we think about the distances $a,b,c?$

How might we think about the distances $a,b,c?$

RisingRook
2020-05-14 15:28:10

trinagle inquality

trinagle inquality

eibc
2020-05-14 15:28:10

triangle innequality!

triangle innequality!

menlo
2020-05-14 15:28:10

use triangle inequality

use triangle inequality

jacobwu
2020-05-14 15:28:10

triangle inequality

triangle inequality

Windigo
2020-05-14 15:28:10

We can use triangle inequalities

We can use triangle inequalities

Arnav.Dagar.sd
2020-05-14 15:28:10

Triangles inequality

Triangles inequality

Mathemagician10
2020-05-14 15:28:10

a + b > c

a + b > c

rrusczyk
2020-05-14 15:28:28

Indeed, we'll need $a+b>c$ and so on from the Triangle Inequality.

Indeed, we'll need $a+b>c$ and so on from the Triangle Inequality.

rrusczyk
2020-05-14 15:28:46

But we're going to need to learn a little more about $a$, $b$, and $c$.

But we're going to need to learn a little more about $a$, $b$, and $c$.

MathJams
2020-05-14 15:28:52

perpendicular distances?

perpendicular distances?

rrusczyk
2020-05-14 15:29:08

Perpendicular distances. Hmmmm.... What might that make us think of?

Perpendicular distances. Hmmmm.... What might that make us think of?

amuthup
2020-05-14 15:29:19

heights of triangles ABP, BCP, CAP

heights of triangles ABP, BCP, CAP

mathgirl199
2020-05-14 15:29:26

triangle heights.

triangle heights.

shalissa
2020-05-14 15:29:29

height of a triangle

height of a triangle

MathWizard38025
2020-05-14 15:29:35

Heights

Heights

PShucks
2020-05-14 15:29:35

heights

heights

douglasubella
2020-05-14 15:29:39

Heights!

Heights!

sixoneeight
2020-05-14 15:29:39

The heihts

The heihts

AbhiWwis
2020-05-14 15:29:39

altitudes

altitudes

ZippityA
2020-05-14 15:29:39

altitudes

altitudes

mathisfun17
2020-05-14 15:29:39

altitudes

altitudes

rrusczyk
2020-05-14 15:29:44

We notice that our distances from $P$ to the sides of $\triangle ABC$ are altitudes of triangles $PAB$, $PBC$, and $PCA$.

We notice that our distances from $P$ to the sides of $\triangle ABC$ are altitudes of triangles $PAB$, $PBC$, and $PCA$.

rrusczyk
2020-05-14 15:29:48

rayfish
2020-05-14 15:30:09

sum of heights must be the same

sum of heights must be the same

rrusczyk
2020-05-14 15:30:16

Bold claim!

Bold claim!

DreamShake34
2020-05-14 15:30:28

i it right?

i it right?

Facejo
2020-05-14 15:30:33

How do we prove?

How do we prove?

rrusczyk
2020-05-14 15:30:39

Let's see!

Let's see!

rrusczyk
2020-05-14 15:30:57

What do altitudes make us think of?

What do altitudes make us think of?

amuthupss
2020-05-14 15:31:14

Areas

Areas

undefined_
2020-05-14 15:31:14

area

area

Math5K
2020-05-14 15:31:14

area

area

Kevin06
2020-05-14 15:31:14

Area!

Area!

Leonard_my_dude
2020-05-14 15:31:14

areas

areas

Chesssaga
2020-05-14 15:31:14

Area

Area

CoolCarsOnTheRun
2020-05-14 15:31:14

areas

areas

green_arrow
2020-05-14 15:31:14

area

area

RankDoughnut255
2020-05-14 15:31:14

area

area

Wesman
2020-05-14 15:31:14

Area

Area

ryanfu2008
2020-05-14 15:31:14

area

area

rrusczyk
2020-05-14 15:31:28

Great; let's talk area.

Great; let's talk area.

Leonard_my_dude
2020-05-14 15:31:51

sum of areas is area of entire triangle so sum of heights is equal to total height

sum of areas is area of entire triangle so sum of heights is equal to total height

IceWolf10
2020-05-14 15:31:51

they have to add up to the same area

they have to add up to the same area

AlphaBetaGammaOmega
2020-05-14 15:31:51

The total area is constant

The total area is constant

Noam2007
2020-05-14 15:31:51

$[APC]+[BPC]+[CPA]=[ABC].$

$[APC]+[BPC]+[CPA]=[ABC].$

I_like_elmo
2020-05-14 15:31:57

ohh, the areas of the traingle will always equal the big traingle, and the bases are the same, therefore, the altidudes have the same sum

ohh, the areas of the traingle will always equal the big traingle, and the bases are the same, therefore, the altidudes have the same sum

rrusczyk
2020-05-14 15:32:09

Very pretty. Let's get through this step by step.

Very pretty. Let's get through this step by step.

rrusczyk
2020-05-14 15:32:13

Letting $[ABC]$ represent the area of triangle $ABC,$ we can split up $ABC$ as $$[ABC] = [PAB] + [PBC] + [PCA].$$

Letting $[ABC]$ represent the area of triangle $ABC,$ we can split up $ABC$ as $$[ABC] = [PAB] + [PBC] + [PCA].$$

rrusczyk
2020-05-14 15:32:23

And how can we write the areas of those smaller triangles?

And how can we write the areas of those smaller triangles?

amuthup
2020-05-14 15:33:17

side length * a/2, side length * b/2, side length * c/2

side length * a/2, side length * b/2, side length * c/2

rocketsri
2020-05-14 15:33:17

11a/2+11b/2+11c/2

11a/2+11b/2+11c/2

Meyzeek_Saveer
2020-05-14 15:33:17

11a +11b+11c all over 2

11a +11b+11c all over 2

greenmathcounts
2020-05-14 15:33:17

11a/2+11b/2+11c/2

11a/2+11b/2+11c/2

AWaltz
2020-05-14 15:33:17

11a/2, 11b/2, and 11c/2

11a/2, 11b/2, and 11c/2

pvmathnerd
2020-05-14 15:33:17

11a/2+11b/2+11c/2

11a/2+11b/2+11c/2

emerald_block
2020-05-14 15:33:17

$\frac{as}{2}, \frac{bs}{2}, \frac{bs}{2}$ where $s$ is the side length

$\frac{as}{2}, \frac{bs}{2}, \frac{bs}{2}$ where $s$ is the side length

rrusczyk
2020-05-14 15:33:26

We use the sides of $ABC$ and the altitudes that we know: \begin{align*}[ABC] &= [PAB]+[PBC]+[PCA]\\

&= \frac{(AB)(c)}{2} + \frac{(BC)(a)}{2} + \frac{(CA)(b)}{2}.\end{align*}

We use the sides of $ABC$ and the altitudes that we know: \begin{align*}[ABC] &= [PAB]+[PBC]+[PCA]\\

&= \frac{(AB)(c)}{2} + \frac{(BC)(a)}{2} + \frac{(CA)(b)}{2}.\end{align*}

rrusczyk
2020-05-14 15:33:34

We know that $AB = AC = BC = 11$. (Pretty convenient that triangle $ABC$ is equilateral!)

We know that $AB = AC = BC = 11$. (Pretty convenient that triangle $ABC$ is equilateral!)

rrusczyk
2020-05-14 15:33:38

So this simplifies to just $[ABC] = \dfrac{11}{2}(a+b+c).$

So this simplifies to just $[ABC] = \dfrac{11}{2}(a+b+c).$

rrusczyk
2020-05-14 15:33:44

Um, how does that help?

Um, how does that help?

snorlax186
2020-05-14 15:34:15

a+b+c is constant

a+b+c is constant

yizhou76
2020-05-14 15:34:15

a+b+c is constant

a+b+c is constant

christopherp
2020-05-14 15:34:15

a+b+c must be constant

a+b+c must be constant

MR_67
2020-05-14 15:34:15

You know the sum of a, b, and c

You know the sum of a, b, and c

Zoobat
2020-05-14 15:34:15

a+b+c is constant

a+b+c is constant

A395134
2020-05-14 15:34:15

$a+b+c$ is constant.

$a+b+c$ is constant.

eibc
2020-05-14 15:34:15

now we know that a+b+c = the height

now we know that a+b+c = the height

rrusczyk
2020-05-14 15:34:35

The area of $ABC$ is constant, so $a+b+c$ must be constant.

The area of $ABC$ is constant, so $a+b+c$ must be constant.

rrusczyk
2020-05-14 15:34:40

Specifically, we note that if $h$ is the height of $\triangle ABC,$ we have $[ABC] = \dfrac{1}{2}(BC)(h) = \dfrac{11}{2}(h).$

Specifically, we note that if $h$ is the height of $\triangle ABC,$ we have $[ABC] = \dfrac{1}{2}(BC)(h) = \dfrac{11}{2}(h).$

MathWizard10
2020-05-14 15:34:59

a+b+c=height

a+b+c=height

jason543
2020-05-14 15:35:05

so h =a+b+c

so h =a+b+c

Tony_Li2007
2020-05-14 15:35:07

a+b+c is height

a+b+c is height

MR_67
2020-05-14 15:35:13

a+b+c is always height

a+b+c is always height

rrusczyk
2020-05-14 15:35:16

So $a+b+c = h,$ the height of $ABC.$

So $a+b+c = h,$ the height of $ABC.$

dc495
2020-05-14 15:35:31

yay we proved it!

yay we proved it!

rrusczyk
2020-05-14 15:35:37

We have indeed!!!

We have indeed!!!

yukrant1
2020-05-14 15:35:44

This is Viviani’s theorem

This is Viviani’s theorem

v4913
2020-05-14 15:35:44

That's Viviani's theorem

That's Viviani's theorem

rrusczyk
2020-05-14 15:35:50

I'm going to take your word for it

I'm going to take your word for it

rrusczyk
2020-05-14 15:35:58

Now what?

Now what?

turtles0120
2020-05-14 15:36:09

but we're not done?

but we're not done?

rrusczyk
2020-05-14 15:36:35

Correct -- we aren't there yet. We need to figure out how many points we can choose that will satisfy our condition in the problem.

Correct -- we aren't there yet. We need to figure out how many points we can choose that will satisfy our condition in the problem.

rrusczyk
2020-05-14 15:36:43

How are we going to do that?

How are we going to do that?

AbhiWwis
2020-05-14 15:36:55

now we look at all the scenarios

now we look at all the scenarios

AbhiWwis
2020-05-14 15:36:55

cant we just count

cant we just count

MR_67
2020-05-14 15:36:55

list out the possibilities

list out the possibilities

rrusczyk
2020-05-14 15:37:09

That will work. In fact, that's how I solved it when I did the problem.

That will work. In fact, that's how I solved it when I did the problem.

rrusczyk
2020-05-14 15:37:16

But Harvey told me that wasn't very sporting.

But Harvey told me that wasn't very sporting.

rrusczyk
2020-05-14 15:37:23

He told me there's a cooler way to finish, but he wouldn't show it to me. Let's see if we can find it together.

He told me there's a cooler way to finish, but he wouldn't show it to me. Let's see if we can find it together.

sdattilo2002
2020-05-14 15:37:45

find the heights that satisfy triangle inequality

find the heights that satisfy triangle inequality

v4913
2020-05-14 15:38:09

The Triangle inequality maybe?

The Triangle inequality maybe?

harmonyguan
2020-05-14 15:38:09

triangle inequality

triangle inequality

cindyzou
2020-05-14 15:38:10

triangle inequality

triangle inequality

matharpmath
2020-05-14 15:38:17

Triangle inequality

Triangle inequality

lrjr24
2020-05-14 15:38:17

triangle inequality

triangle inequality

rrusczyk
2020-05-14 15:38:19

Oh yeah, we were talking about the Triangle Inequality a while back.

Oh yeah, we were talking about the Triangle Inequality a while back.

rrusczyk
2020-05-14 15:38:43

The Triangle Inequality tells us that we must have $$\begin{align*} a &< b+c\\ b &< a+c\\ c &< a+b\end{align*}$$in order for $a,b,c$ to be sides of a triangle.

The Triangle Inequality tells us that we must have $$\begin{align*} a &< b+c\\ b &< a+c\\ c &< a+b\end{align*}$$in order for $a,b,c$ to be sides of a triangle.

doglover07
2020-05-14 15:39:22

we also know that a+b+c=h

we also know that a+b+c=h

christopherp
2020-05-14 15:39:22

a+b+c=h

a+b+c=h

jacobwu
2020-05-14 15:39:22

we alredady know that $a+b+c=h$

we alredady know that $a+b+c=h$

Chesssaga
2020-05-14 15:39:32

But it's easier to do with heights right?

But it's easier to do with heights right?

rrusczyk
2020-05-14 15:40:08

Good idea -- we need to relate this to that interesting $a+b+c=h$ thing we just learned. What does this do to our Triangle Inequalities?

Good idea -- we need to relate this to that interesting $a+b+c=h$ thing we just learned. What does this do to our Triangle Inequalities?

Mathworker2705
2020-05-14 15:40:51

We can substitute!!!

We can substitute!!!

mmjguitar
2020-05-14 15:40:51

h-a=b+c

h-a=b+c

Jomo
2020-05-14 15:40:51

h-a>a

h-a>a

nikenissan
2020-05-14 15:41:00

2a < h, 2b < h, and 2c < h

2a < h, 2b < h, and 2c < h

rrusczyk
2020-05-14 15:41:10

We can replace $b+c$ with $h-a$, and similarly in the other two inequalities.

We can replace $b+c$ with $h-a$, and similarly in the other two inequalities.

rrusczyk
2020-05-14 15:41:11

This gives $$\begin{align*} a &< h-a\\ b &< h-b\\ c &< h-c\end{align*}$$

This gives $$\begin{align*} a &< h-a\\ b &< h-b\\ c &< h-c\end{align*}$$

rrusczyk
2020-05-14 15:41:14

And this becomes $2a < h$ and $2b < h$ and $2c < h$.

And this becomes $2a < h$ and $2b < h$ and $2c < h$.

rrusczyk
2020-05-14 15:41:18

So for $a,b,c$ to be the sides of a triangle, they must all be less than $\dfrac{h}{2}$.

So for $a,b,c$ to be the sides of a triangle, they must all be less than $\dfrac{h}{2}$.

DaBobWhoLikeMath1234
2020-05-14 15:41:35

We've limited the possibilities!

We've limited the possibilities!

RisingRook
2020-05-14 15:41:35

that narrows it down a lot

that narrows it down a lot

rrusczyk
2020-05-14 15:41:47

Yes we have!

Yes we have!

rrusczyk
2020-05-14 15:41:57

What are the possibilities that work here?

What are the possibilities that work here?

ISolveLikeAGirl
2020-05-14 15:42:00

so then count the points that work

so then count the points that work

rrusczyk
2020-05-14 15:42:15

Yes, let's count 'em. Where are those points that work?

Yes, let's count 'em. Where are those points that work?

donguri
2020-05-14 15:42:50

so its in the middle quarter of the triangle

so its in the middle quarter of the triangle

Jomo
2020-05-14 15:42:50

We can draw boundaries!

We can draw boundaries!

Jomo
2020-05-14 15:42:50

It can't be above the halfline in each direction!!

It can't be above the halfline in each direction!!

CHIPPER33
2020-05-14 15:42:50

the inner triangle

the inner triangle

hi13
2020-05-14 15:42:50

draw a triangle to box in the possibilities

draw a triangle to box in the possibilities

donguri
2020-05-14 15:42:50

the middle quarter of the triangle

the middle quarter of the triangle

ThatRichDeng
2020-05-14 15:42:50

if you draw all the midlines of the triangle the points in the middle region

if you draw all the midlines of the triangle the points in the middle region

rrusczyk
2020-05-14 15:42:54

We draw the three segments connecting the midpoints of the sides:

We draw the three segments connecting the midpoints of the sides:

rrusczyk
2020-05-14 15:42:55

KevinW12345
2020-05-14 15:43:18

Somewhere around the center

Somewhere around the center

mmjguitar
2020-05-14 15:43:18

around the center

around the center

fasterthanlight
2020-05-14 15:43:18

In the center

In the center

v4913
2020-05-14 15:43:18

Only in the very middle

Only in the very middle

Maths4J
2020-05-14 15:43:18

Divide it into 4 congruent equilateral triangles and the center one works (inside it)

Divide it into 4 congruent equilateral triangles and the center one works (inside it)

Chesssaga
2020-05-14 15:43:32

Medial triangle!

Medial triangle!

rrusczyk
2020-05-14 15:43:37

Any point inside this smaller triangle will be less than $\dfrac{h}{2}$ from every side of the big triangle.

Any point inside this smaller triangle will be less than $\dfrac{h}{2}$ from every side of the big triangle.

rrusczyk
2020-05-14 15:43:51

(And yes, we call that little triangle in the middle the "Medial triangle").

(And yes, we call that little triangle in the middle the "Medial triangle").

rrusczyk
2020-05-14 15:43:54

I think we found Harvey's solution!

I think we found Harvey's solution!

KLBBC
2020-05-14 15:44:12

SO all the points in that region, which is 15??

SO all the points in that region, which is 15??

OlympusHero
2020-05-14 15:44:12

5+4+3+2+1=15

5+4+3+2+1=15

T0-1h3MathMug360
2020-05-14 15:44:12

so its... 1+2+3+4+5=15

so its... 1+2+3+4+5=15

MathWizard10
2020-05-14 15:44:12

that is 15 points

that is 15 points

RedFireTruck
2020-05-14 15:44:12

15

15

Wizkid8402
2020-05-14 15:44:12

$15$

$15$

sixoneeight
2020-05-14 15:44:12

there are 15 "good" points

there are 15 "good" points

universeking
2020-05-14 15:44:12

15 points work!

15 points work!

CT17
2020-05-14 15:44:12

So there are $\frac{5\cdot 6}{2}=15$ points that work!

So there are $\frac{5\cdot 6}{2}=15$ points that work!

rrusczyk
2020-05-14 15:44:25

Counting up the rows, we see that $5+4+3+2+1 = 15$ lattice points are inside that little triangle.

Counting up the rows, we see that $5+4+3+2+1 = 15$ lattice points are inside that little triangle.

rrusczyk
2020-05-14 15:44:41

So our answer is....

So our answer is....

rponda
2020-05-14 15:44:56

1/3

1/3

MathematicianFlippityboii
2020-05-14 15:45:00

15/45=1/3!

15/45=1/3!

BakedPotato66
2020-05-14 15:45:00

15/45= 1/3

15/45= 1/3

yogurt12345
2020-05-14 15:45:00

15/45 is 1/3

15/45 is 1/3

universeking
2020-05-14 15:45:00

probability is 15/45=1/3

probability is 15/45=1/3

yoonsunam
2020-05-14 15:45:00

15/45=1/3

15/45=1/3

Shravi
2020-05-14 15:45:00

15/45=1/3

15/45=1/3

btc433
2020-05-14 15:45:00

The probability is $\frac{15}{45}=\frac13$

The probability is $\frac{15}{45}=\frac13$

bobthegod78
2020-05-14 15:45:00

possible /total 15/45 =1/3 !!!

possible /total 15/45 =1/3 !!!

MathAcademyIsFun
2020-05-14 15:45:00

15/45 = 1/3

15/45 = 1/3

rrusczyk
2020-05-14 15:45:07

The probability is $\dfrac{\text{points that work}}{\text{total points}} = \dfrac{15}{45} = \boxed{\dfrac13}.$

The probability is $\dfrac{\text{points that work}}{\text{total points}} = \dfrac{15}{45} = \boxed{\dfrac13}.$

rrusczyk
2020-05-14 15:45:19

Phew. One down.

Phew. One down.

mathgirl199
2020-05-14 15:45:27

Wow, that was actually pretty quick.

Wow, that was actually pretty quick.

rrusczyk
2020-05-14 15:45:30

rrusczyk
2020-05-14 15:45:42

All right, a reminder: there are over 1300 of you here!

All right, a reminder: there are over 1300 of you here!

rrusczyk
2020-05-14 15:46:13

So, we aren't going to be able to react to all of your questions or comments today

So, we aren't going to be able to react to all of your questions or comments today

rrusczyk
2020-05-14 15:46:22

Our usual classes are much, much smaller!!!

Our usual classes are much, much smaller!!!

jabuticaba149
2020-05-14 15:46:33

That helped a lot thank you!

That helped a lot thank you!

math12345678
2020-05-14 15:46:33

wasn't as hard as I think

wasn't as hard as I think

jason543
2020-05-14 15:46:33

yay!!!!

yay!!!!

Anniboy
2020-05-14 15:46:33

woohoo! Nice job! We solved it!

woohoo! Nice job! We solved it!

Duckyduckcheese
2020-05-14 15:46:33

YAY

YAY

mohanty
2020-05-14 15:46:37

Next Question!

Next Question!

CosmoMonkhouse
2020-05-14 15:46:37

many to go

many to go

DarthMaul
2020-05-14 15:46:39

whats next

whats next

rrusczyk
2020-05-14 15:46:50

That's one of my favorite lines...

That's one of my favorite lines...

rrusczyk
2020-05-14 15:46:51

**Sprint #26:**A circle is tangent to the positive $x$-axis at $x = 3.$ It passes through the distinct points $(6, 6)$ and $(p, p).$ What is the value of $p?$ Express your answer as a common fraction.
rrusczyk
2020-05-14 15:47:01

What do we start with here?

What do we start with here?

PerfectMath
2020-05-14 15:47:21

make a graph??

make a graph??

rondoallaturca
2020-05-14 15:47:21

graph what we can

graph what we can

doglover07
2020-05-14 15:47:21

draw a picture

draw a picture

coldcrazylogic
2020-05-14 15:47:21

draw?

draw?

flissyquokka17
2020-05-14 15:47:21

draw a diagram, right?

draw a diagram, right?

PoisonApple
2020-05-14 15:47:25

drawing it

drawing it

rrusczyk
2020-05-14 15:47:38

I can't get anywhere without a picture. I'm not Harvey.

I can't get anywhere without a picture. I'm not Harvey.

rrusczyk
2020-05-14 15:47:40

Let's sketch a picture of what's going on.

Let's sketch a picture of what's going on.

rrusczyk
2020-05-14 15:47:42

greenmathcounts
2020-05-14 15:48:14

center is (3,y)

center is (3,y)

asbodke
2020-05-14 15:48:14

The center of the circle lies on x=3

The center of the circle lies on x=3

dineshs
2020-05-14 15:48:14

How do you assume that p<3

How do you assume that p<3

rrusczyk
2020-05-14 15:49:23

Indeed the center will be on the vertical line through $x=3$. This gives us the answer to the question dineshs asks (which is very much a good, non-obvious question!)

Indeed the center will be on the vertical line through $x=3$. This gives us the answer to the question dineshs asks (which is very much a good, non-obvious question!)

ccu700037
2020-05-14 15:49:37

p must be less than 3 because it must fall on y=x, and a line can only cross a circle at 2 points

p must be less than 3 because it must fall on y=x, and a line can only cross a circle at 2 points

DZL1
2020-05-14 15:49:37

and p,p and 6,6 are both on the line y=x... oh

and p,p and 6,6 are both on the line y=x... oh

rrusczyk
2020-05-14 15:50:02

Exactly -- $x=y$ hits our circle at two points, one on either side of the line $x=3$.

Exactly -- $x=y$ hits our circle at two points, one on either side of the line $x=3$.

rrusczyk
2020-05-14 15:50:17

So, now we have:

So, now we have:

rrusczyk
2020-05-14 15:50:23

rrusczyk
2020-05-14 15:51:48

Yikes -- I skipped a bit of step there. Why is the $y$-coordinate of the center of the circle equal to our radius?

Yikes -- I skipped a bit of step there. Why is the $y$-coordinate of the center of the circle equal to our radius?

rrusczyk
2020-05-14 15:51:56

(Which I've clearly called $r$)

(Which I've clearly called $r$)

AwesomeLife_Math
2020-05-14 15:52:47

The distance from (3,r) to (6,6) must be equal to r.

The distance from (3,r) to (6,6) must be equal to r.

nihao4112
2020-05-14 15:52:47

Because of tangency

Because of tangency

LANSH
2020-05-14 15:52:47

TANGENT

TANGENT

Puffer13
2020-05-14 15:52:47

it is tangent to the x-axis

it is tangent to the x-axis

mathdolphin123
2020-05-14 15:52:47

The circle is tangent

The circle is tangent

tigerzhang
2020-05-14 15:52:49

it is tangent to the x-axis

it is tangent to the x-axis

BakedPotato66
2020-05-14 15:52:51

it's tangent to the x axis

it's tangent to the x axis

rrusczyk
2020-05-14 15:52:52

Ah right, the circle is tangent to the $x$-axis, so the segment from the center to $(3,0)$ is a radius.

Ah right, the circle is tangent to the $x$-axis, so the segment from the center to $(3,0)$ is a radius.

rrusczyk
2020-05-14 15:53:08

What's next?

What's next?

Zoobat
2020-05-14 15:53:18

so we use the distance formula to find r?

so we use the distance formula to find r?

rrusczyk
2020-05-14 15:53:20

How?

How?

jacobwu
2020-05-14 15:53:33

we can come up with an equation?

we can come up with an equation?

rrusczyk
2020-05-14 15:53:40

Good plan. How do we get the equation?

Good plan. How do we get the equation?

BBC0506
2020-05-14 15:53:48

Use the distance formula to find the distance from (3,r) to (6,6)

Use the distance formula to find the distance from (3,r) to (6,6)

superagh
2020-05-14 15:53:48

distance to (6, 6) is equal to r

distance to (6, 6) is equal to r

Maths4J
2020-05-14 15:53:48

Distance formula from (3, r) to (6, 6), then solve for r.

Distance formula from (3, r) to (6, 6), then solve for r.

scinderella220
2020-05-14 15:53:51

r=the Distance from (3,r) to (6,6)

r=the Distance from (3,r) to (6,6)

vsamc
2020-05-14 15:53:53

distance from (3,r) to (6,6) must be equal to r

distance from (3,r) to (6,6) must be equal to r

rrusczyk
2020-05-14 15:54:00

The center must be $r$ away from $(6,6).$

The center must be $r$ away from $(6,6).$

rrusczyk
2020-05-14 15:54:05

What equation does this give us?

What equation does this give us?

youcandomathnow
2020-05-14 15:54:29

(6-3)^2 + (6-r)^2 = r^2

(6-3)^2 + (6-r)^2 = r^2

asbodke
2020-05-14 15:54:29

r^2 = (6-r)^2 + 9

r^2 = (6-r)^2 + 9

Jomo
2020-05-14 15:54:29

$r^2=3^2+(6-r)^2$

$r^2=3^2+(6-r)^2$

IMadeYouReadThis
2020-05-14 15:54:29

$r^2=3^2+(6-r)^2$

$r^2=3^2+(6-r)^2$

Speedstorm
2020-05-14 15:54:39

$(6-3)^2+(6-r)^2=r^2$

$(6-3)^2+(6-r)^2=r^2$

christopherp
2020-05-14 15:54:43

(6-3)^2 + (6-r)^2 = r^2

(6-3)^2 + (6-r)^2 = r^2

rrusczyk
2020-05-14 15:54:47

We must have \[(6-3)^2 + (6-r)^2 = r^2.\]

We must have \[(6-3)^2 + (6-r)^2 = r^2.\]

rrusczyk
2020-05-14 15:54:58

Time for some algebra.

Time for some algebra.

ZippityA
2020-05-14 15:55:04

Expand.

Expand.

jacobwu
2020-05-14 15:55:09

expand it

expand it

Piano_Man123
2020-05-14 15:55:21

Expand

Expand

ttmmrryy
2020-05-14 15:55:22

expand

expand

rrusczyk
2020-05-14 15:55:25

Keep going!

Keep going!

Jomo
2020-05-14 15:55:43

$r^2=r^2-12r+45$

$r^2=r^2-12r+45$

vsurya
2020-05-14 15:55:51

9+36-12r+r^2=r^2

9+36-12r+r^2=r^2

menlo
2020-05-14 15:55:52

9+36 - 12r+r^2=r^2

9+36 - 12r+r^2=r^2

winterrain01
2020-05-14 15:55:54

9+36-12r+r^2=r^2

9+36-12r+r^2=r^2

rrusczyk
2020-05-14 15:55:59

Simplifying the left-hand side gives $r^2 - 12r + 45 = r^2.$

Simplifying the left-hand side gives $r^2 - 12r + 45 = r^2.$

xMidnightFirex
2020-05-14 15:56:22

$r = \frac{15}4$

$r = \frac{15}4$

bobthegod78
2020-05-14 15:56:22

12r=45

12r=45

Scipow
2020-05-14 15:56:22

r = 15/4

r = 15/4

hi13
2020-05-14 15:56:22

12r=45

12r=45

bobthegod78
2020-05-14 15:56:22

12r=45

12r=45

KLBBC
2020-05-14 15:56:22

r=15/4

r=15/4

yizhou76
2020-05-14 15:56:22

r=15/4

r=15/4

Raiders26
2020-05-14 15:56:26

15/4=r

15/4=r

Mathqueen2018
2020-05-14 15:56:26

r = 15/4

r = 15/4

rrusczyk
2020-05-14 15:56:30

The $r^2$'s cancel, and what we're left with gives $r= \dfrac{45}{12} = \dfrac{15}{4}.$

The $r^2$'s cancel, and what we're left with gives $r= \dfrac{45}{12} = \dfrac{15}{4}.$

rrusczyk
2020-05-14 15:56:32

rrusczyk
2020-05-14 15:56:37

Now what?

Now what?

Bryguy
2020-05-14 15:57:06

use distance formula again

use distance formula again

Chesssaga
2020-05-14 15:57:06

Now use distance formula to calculate p

Now use distance formula to calculate p

walrus987
2020-05-14 15:57:06

distance formula with p?

distance formula with p?

mathgirl199
2020-05-14 15:57:09

Now use distance formula for (3,15/4) and (p,p)

Now use distance formula for (3,15/4) and (p,p)

rrusczyk
2020-05-14 15:57:11

Often when a strategy works once in a problem, it will work again.

Often when a strategy works once in a problem, it will work again.

AwesomeLife_Math
2020-05-14 15:57:18

Distance formula on (p,p).

Distance formula on (p,p).

ThatRichDeng
2020-05-14 15:57:18

distance formula again

distance formula again

math12345678
2020-05-14 15:57:20

do the same thing to (p,p)

do the same thing to (p,p)

rrusczyk
2020-05-14 15:57:25

The distance from $(p,p)$ to the center must be $r$ too!

The distance from $(p,p)$ to the center must be $r$ too!

rrusczyk
2020-05-14 15:57:26

And we can write an equation for that, too.

And we can write an equation for that, too.

rrusczyk
2020-05-14 15:57:28

\[(p-3)^2+\left(p-\frac{15}{4}\right)^2 = \left(\frac{15}{4}\right)^2.\]

\[(p-3)^2+\left(p-\frac{15}{4}\right)^2 = \left(\frac{15}{4}\right)^2.\]

rrusczyk
2020-05-14 15:57:46

What might you do first on this to make it easier to deal with?

What might you do first on this to make it easier to deal with?

Alculator11
2020-05-14 15:58:11

Multiply by 16

Multiply by 16

MathJams
2020-05-14 15:58:11

multiply both sides by 16?

multiply both sides by 16?

Bloppity
2020-05-14 15:58:11

Get rid of fractions?

Get rid of fractions?

Kevin06
2020-05-14 15:58:11

Multiply by 16?

Multiply by 16?

winterrain01
2020-05-14 15:58:11

get rid of the fraction?

get rid of the fraction?

TheEpicCarrot7
2020-05-14 15:58:11

Multiply by 16!

Multiply by 16!

Puffer13
2020-05-14 15:58:11

multiply bot hsides by 16

multiply bot hsides by 16

rrusczyk
2020-05-14 15:58:16

I'd start by multiplying both sides by $16$, to get rid of the fractions.

I'd start by multiplying both sides by $16$, to get rid of the fractions.

rrusczyk
2020-05-14 15:58:17

We get \[16(p-3)^2 + (4p - 15)^2 = 225.\]

We get \[16(p-3)^2 + (4p - 15)^2 = 225.\]

rrusczyk
2020-05-14 15:58:26

The rest is algebra.

The rest is algebra.

TMSmathcounts737
2020-05-14 15:58:42

Now expand.

Now expand.

Cozzmo
2020-05-14 15:58:42

expand that

expand that

cj13609517288
2020-05-14 15:58:42

expand

expand

bobthefam
2020-05-14 15:58:42

Expand!

Expand!

ElNoraa
2020-05-14 15:58:42

expand?

expand?

G.G.Otto
2020-05-14 15:58:42

now expand

now expand

mathgirl199
2020-05-14 15:58:42

Now we can expand!

Now we can expand!

Mathematician1010
2020-05-14 15:58:42

now expand and simplify

now expand and simplify

universeking
2020-05-14 15:58:42

expand

expand

Asterlan
2020-05-14 15:58:42

Then expand

Then expand

rrusczyk
2020-05-14 15:58:48

Now, we can expand the left side and simplify, to get \[

32p^2 - 216p +144=0.\]

Now, we can expand the left side and simplify, to get \[

32p^2 - 216p +144=0.\]

Zoobat
2020-05-14 15:58:56

4p^2 - 27p + 18 = 0

4p^2 - 27p + 18 = 0

rrusczyk
2020-05-14 15:59:24

Dividing both sides by $8$ gives $4p^2 - 27p +18=0.$

Dividing both sides by $8$ gives $4p^2 - 27p +18=0.$

asbodke
2020-05-14 15:59:33

(p-6)(4p-3) = 0

(p-6)(4p-3) = 0

huela
2020-05-14 15:59:53

(4p-3)(p-6)=0

(4p-3)(p-6)=0

rrusczyk
2020-05-14 15:59:54

Factoring gives $(4p-3)(p-6)=0.$

Factoring gives $(4p-3)(p-6)=0.$

rrusczyk
2020-05-14 16:00:00

Uh-oh. Two solutions.

Uh-oh. Two solutions.

doglover07
2020-05-14 16:00:26

and p is not 6

and p is not 6

Chesssaga
2020-05-14 16:00:26

p = 3/4 since we proved it is not greater than 3

p = 3/4 since we proved it is not greater than 3

Bryguy
2020-05-14 16:00:26

p cannot equal 6

p cannot equal 6

winterrain01
2020-05-14 16:00:27

p must be the smaller solution

p must be the smaller solution

rrusczyk
2020-05-14 16:00:59

Ah, that's right. We want the point that's not $(6,6)$. So $p$ is...

Ah, that's right. We want the point that's not $(6,6)$. So $p$ is...

PV123
2020-05-14 16:01:19

3/4

3/4

mark888
2020-05-14 16:01:19

3/4!

3/4!

hi13
2020-05-14 16:01:19

3/4

3/4

alphaKITTY37
2020-05-14 16:01:19

3/ 4

3/ 4

azhang1202
2020-05-14 16:01:19

3/4

3/4

RJ5303707
2020-05-14 16:01:19

$3/4$

$3/4$

BananaWatch5223
2020-05-14 16:01:19

3/4

3/4

ChessSavage
2020-05-14 16:01:19

3/4

3/4

colorpencilz
2020-05-14 16:01:19

3/4

3/4

MathWizard10
2020-05-14 16:01:19

3/4

3/4

bonami2018
2020-05-14 16:01:19

3/4

3/4

kingdragonshadow
2020-05-14 16:01:19

3/4

3/4

christopherp
2020-05-14 16:01:19

3/4

3/4

arisettyjr
2020-05-14 16:01:19

3/4

3/4

WAWATHEGOAT
2020-05-14 16:01:19

3/4

3/4

iam_awesome
2020-05-14 16:01:23

$$3/4$$

$$3/4$$

joseph2718
2020-05-14 16:01:23

3/4

3/4

robot317
2020-05-14 16:01:23

p is 3/4

p is 3/4

rrusczyk
2020-05-14 16:01:28

So we take the solution $p = \boxed{\dfrac34}.$

So we take the solution $p = \boxed{\dfrac34}.$

TheEpicCarrot7
2020-05-14 16:01:50

Yay we solved it!

Yay we solved it!

scinderella220
2020-05-14 16:01:50

YAY!

YAY!

Mistyketchum28
2020-05-14 16:01:50

Yay!

Yay!

tacos4life123
2020-05-14 16:01:50

yeah

yeah

Terribleteeth
2020-05-14 16:01:50

yay! we did it

yay! we did it

CHIPPER33
2020-05-14 16:01:50

YAY! NEXT

YAY! NEXT

Raiders26
2020-05-14 16:01:50

Yay!!!

Yay!!!

sahila
2020-05-14 16:01:50

yay

yay

MathWizard38025
2020-05-14 16:01:50

thats a good solution

thats a good solution

Mistyketchum28
2020-05-14 16:01:54

That was a good one!

That was a good one!

SandyBeach11
2020-05-14 16:01:54

YAY WE DID IT

YAY WE DID IT

sahanapotu
2020-05-14 16:01:57

now it amkes sense

now it amkes sense

rrusczyk
2020-05-14 16:02:10

Great!

Great!

punkinpiday
2020-05-14 16:02:14

Two down, four to go!

Two down, four to go!

rrusczyk
2020-05-14 16:02:25

Here's the next one:

Here's the next one:

rrusczyk
2020-05-14 16:02:25

**Sprint #27**: How many of the first $2019$ positive integers have no odd single-digit prime factors?
v4913
2020-05-14 16:02:47

THIS WAS AN AWESOME PROBLEM XD

THIS WAS AN AWESOME PROBLEM XD

rrusczyk
2020-05-14 16:02:52

Glad you approve

Glad you approve

Equinox8
2020-05-14 16:03:18

complementary counting?

complementary counting?

srim1027
2020-05-14 16:03:18

use complementary counting?

use complementary counting?

ryanfu2008
2020-05-14 16:03:18

complementary counting???

complementary counting???

awsomek
2020-05-14 16:03:18

complementary count?

complementary count?

rrusczyk
2020-05-14 16:03:32

For those of you who don't know this term:

For those of you who don't know this term:

MR_67
2020-05-14 16:03:34

do we count the "not"

do we count the "not"

rrusczyk
2020-05-14 16:03:50

"Complementary counting" is a fancy way to say "count the thing you don't want".

"Complementary counting" is a fancy way to say "count the thing you don't want".

rrusczyk
2020-05-14 16:04:04

What's a clue in this problem that we might want to count the things we don't want?

What's a clue in this problem that we might want to count the things we don't want?

harmonyguan
2020-05-14 16:04:33

"no"

"no"

RisingRook
2020-05-14 16:04:33

the word no

the word no

panda2018
2020-05-14 16:04:33

'no'

'no'

Monkey287
2020-05-14 16:04:33

have NO

have NO

green_alligator
2020-05-14 16:04:33

it says "have no"

it says "have no"

HappySpring
2020-05-14 16:04:33

the "no"

the "no"

doglover07
2020-05-14 16:04:33

"no"

"no"

DottedCaculator
2020-05-14 16:04:33

NO

NO

AlphaBetaGammaOmega
2020-05-14 16:04:33

the word "no"

the word "no"

aops_band
2020-05-14 16:04:40

no

no

iam_awesome
2020-05-14 16:04:40

the word "no"

the word "no"

razormouth
2020-05-14 16:04:40

the word "no"

the word "no"

Math5K
2020-05-14 16:04:40

no

no

MyNameIsJeffZF
2020-05-14 16:04:40

NO odd ...

NO odd ...

superagh
2020-05-14 16:04:40

"no"

"no"

rrusczyk
2020-05-14 16:04:58

"No" and "not" are clues. There's also just the number of cases to think about.

"No" and "not" are clues. There's also just the number of cases to think about.

rrusczyk
2020-05-14 16:05:05

What are the factors we have to avoid?

What are the factors we have to avoid?

asbodke
2020-05-14 16:05:18

3,5,7

3,5,7

jupiter314
2020-05-14 16:05:18

3,5 and 7

3,5 and 7

fasterthanlight
2020-05-14 16:05:18

3, 5, and 7

3, 5, and 7

NHatAOPS
2020-05-14 16:05:18

3 5 and 7

3 5 and 7

sv739
2020-05-14 16:05:18

3,5,7

3,5,7

Leonard_my_dude
2020-05-14 16:05:18

3,5,7

3,5,7

n3v3rmor3
2020-05-14 16:05:18

3, 5, 7

3, 5, 7

Jerry_Guo
2020-05-14 16:05:18

3,5,7

3,5,7

AMC_Kid
2020-05-14 16:05:18

3, 5, 7

3, 5, 7

math_piggy
2020-05-14 16:05:18

3, 5, and 7

3, 5, and 7

os31415
2020-05-14 16:05:18

3, 5, and 7

3, 5, and 7

yayatheduck
2020-05-14 16:05:18

3, 5, 7

3, 5, 7

snorlax186
2020-05-14 16:05:18

3,5, and 7

3,5, and 7

SkywalkerAUV
2020-05-14 16:05:18

3, 5, and 7

3, 5, and 7

IMadeYouReadThis
2020-05-14 16:05:18

$3, 5, 7$

$3, 5, 7$

tenebrine
2020-05-14 16:05:21

3, 5, 7

3, 5, 7

Mathqueen2018
2020-05-14 16:05:21

3, 5, 7

3, 5, 7

Speedstorm
2020-05-14 16:05:22

3,5 and 7

3,5 and 7

maxxie2019
2020-05-14 16:05:24

3, 5, 7

3, 5, 7

C0atimundi
2020-05-14 16:05:24

3, 5, 7

3, 5, 7

Minibangs
2020-05-14 16:05:24

3, 5, and 7

3, 5, and 7

rrusczyk
2020-05-14 16:05:28

We need to avoid the odd single-digit primes, which are $3$, $5$, and $7.$

We need to avoid the odd single-digit primes, which are $3$, $5$, and $7.$

rrusczyk
2020-05-14 16:05:39

That leave us with all the numbers that only have factors $2, 11, 13, 17,$ etc., but that's a lot of primes to keep track of! I don't even know how many there are!

That leave us with all the numbers that only have factors $2, 11, 13, 17,$ etc., but that's a lot of primes to keep track of! I don't even know how many there are!

rrusczyk
2020-05-14 16:06:02

So, the stuff we want to count covers a TON of cases. The stuff we don't want just covers 3 primes.

So, the stuff we want to count covers a TON of cases. The stuff we don't want just covers 3 primes.

sixoneeight
2020-05-14 16:06:11

Subtract the cases you don't want from the total

Subtract the cases you don't want from the total

rrusczyk
2020-05-14 16:06:27

That's the plan. Let's count how many of the first $2019$ positive integers have $3, 5,$ or $7$ as a factor. This counts the numbers we

That's the plan. Let's count how many of the first $2019$ positive integers have $3, 5,$ or $7$ as a factor. This counts the numbers we

__don't__want.
rrusczyk
2020-05-14 16:06:33

Then we'll subtract this count from $2019$ to get our answer: the count of the numbers we do want.

Then we'll subtract this count from $2019$ to get our answer: the count of the numbers we do want.

rrusczyk
2020-05-14 16:06:35

How many of them have $3$ as a factor?

How many of them have $3$ as a factor?

smartatmath
2020-05-14 16:06:57

673

673

The_Better_Samuel
2020-05-14 16:06:57

673

673

aie8920
2020-05-14 16:06:57

673

673

Ninja11
2020-05-14 16:06:57

673

673

Lux1
2020-05-14 16:06:57

there are 673

there are 673

Jomo
2020-05-14 16:06:57

673

673

nye
2020-05-14 16:06:57

673

673

HappySpring
2020-05-14 16:06:57

673

673

AwesomeLife_Math
2020-05-14 16:06:57

673.

673.

bot101
2020-05-14 16:06:57

673

673

lihao_david
2020-05-14 16:06:57

673

673

dhaops
2020-05-14 16:06:57

673

673

Awesome_360
2020-05-14 16:06:57

673

673

krishgarg
2020-05-14 16:06:57

673

673

Meyzeek_Saveer
2020-05-14 16:06:57

673

673

Happytwin
2020-05-14 16:06:57

673

673

Quaoar
2020-05-14 16:06:57

There are 673 multiples of 3

There are 673 multiples of 3

IAmTheHazard
2020-05-14 16:07:07

Floor(2019/3)=673

Floor(2019/3)=673

rrusczyk
2020-05-14 16:07:10

We compute that $2019 \div 3 = 673.$

We compute that $2019 \div 3 = 673.$

rrusczyk
2020-05-14 16:07:12

So there are $673$ numbers in our set that are a multiple of $3.$

So there are $673$ numbers in our set that are a multiple of $3.$

rrusczyk
2020-05-14 16:07:13

How many of them have $5$ as a factor?

How many of them have $5$ as a factor?

KOYO
2020-05-14 16:07:43

403

403

bluemathcounts
2020-05-14 16:07:43

403

403

justinzhang404
2020-05-14 16:07:43

403

403

SparklyFlowers
2020-05-14 16:07:43

403

403

nose
2020-05-14 16:07:43

403

403

robot317
2020-05-14 16:07:43

403

403

krithikrokcs
2020-05-14 16:07:43

403

403

cat_maniac_8
2020-05-14 16:07:43

403

403

melodyzhao
2020-05-14 16:07:43

403

403

twinbrian
2020-05-14 16:07:43

$403$

$403$

scoutskylar
2020-05-14 16:07:43

403

403

rayfish
2020-05-14 16:07:43

$\lfloor \frac{2019}{5}\rfloor = 403$

$\lfloor \frac{2019}{5}\rfloor = 403$

Pokemon2
2020-05-14 16:07:43

floor(2019/5)=403

floor(2019/5)=403

tejasgandi
2020-05-14 16:07:43

403

403

palindrome868
2020-05-14 16:07:43

403

403

dineshs
2020-05-14 16:07:43

2019/5 = 403.8, so 403

2019/5 = 403.8, so 403

rrusczyk
2020-05-14 16:07:48

We compute that $2019 \div 5 = 403\frac45.$

We compute that $2019 \div 5 = 403\frac45.$

rrusczyk
2020-05-14 16:07:57

How many of them have $7$ as a factor?

How many of them have $7$ as a factor?

MathMaster029
2020-05-14 16:08:18

288

288

yizhou76
2020-05-14 16:08:18

288

288

hungrypig
2020-05-14 16:08:18

288

288

TheEpicCarrot7
2020-05-14 16:08:18

288

288

ElNoraa
2020-05-14 16:08:18

288

288

twinklebrownie
2020-05-14 16:08:18

288

288

Math5K
2020-05-14 16:08:18

288

288

PureSwag
2020-05-14 16:08:18

$\lfloor \frac{2019}{7} \rfloor = 288$

$\lfloor \frac{2019}{7} \rfloor = 288$

WAWATHEGOAT
2020-05-14 16:08:18

288

288

bobthefam
2020-05-14 16:08:18

288

288

douglasubella
2020-05-14 16:08:18

288

288

hxz913
2020-05-14 16:08:18

288

288

NonOxyCrustacean
2020-05-14 16:08:18

288

288

Awesome_360
2020-05-14 16:08:18

288

288

gordonhero
2020-05-14 16:08:18

288

288

enkitty
2020-05-14 16:08:18

Floor(2019/7)=288

Floor(2019/7)=288

rrusczyk
2020-05-14 16:08:21

We compute that $2019 \div 7 = 288\frac37.$

We compute that $2019 \div 7 = 288\frac37.$

rrusczyk
2020-05-14 16:08:22

So there are $288$ numbers in our set that are a multiple of $7.$

So there are $288$ numbers in our set that are a multiple of $7.$

rrusczyk
2020-05-14 16:08:27

It looks like this gives us $673+403+288 = 1364$ numbers total that we don't want. Is that right?

It looks like this gives us $673+403+288 = 1364$ numbers total that we don't want. Is that right?

pog
2020-05-14 16:08:40

It's just a venn diagram question really

It's just a venn diagram question really

rrusczyk
2020-05-14 16:09:06

Oh, interesting comment --- Venn diagrams have...

Oh, interesting comment --- Venn diagrams have...

dws1188
2020-05-14 16:09:14

overlap

overlap

RedFireTruck
2020-05-14 16:09:14

OVERLAPS

OVERLAPS

rrusczyk
2020-05-14 16:09:17

Overlaps.

Overlaps.

joseph2718
2020-05-14 16:09:32

No, we're overcounting

No, we're overcounting

rayfish
2020-05-14 16:09:32

No, we are overcounting

No, we are overcounting

NHatAOPS
2020-05-14 16:09:32

WE OVERCOUNTED!

WE OVERCOUNTED!

Raiders26
2020-05-14 16:09:32

overcounting

overcounting

MathJams
2020-05-14 16:09:32

no we overcounted

no we overcounted

vsamc
2020-05-14 16:09:32

no but we are overcounting

no but we are overcounting

Juno
2020-05-14 16:09:32

WE OVERCOUNTED

WE OVERCOUNTED

MathyAOP
2020-05-14 16:09:32

Overcounting

Overcounting

knoxboy
2020-05-14 16:09:32

some numbers overlap

some numbers overlap

aopscasanjose
2020-05-14 16:09:32

no you overcounted

no you overcounted

rrusczyk
2020-05-14 16:09:43

We overcounted! We counted some numbers twice.

We overcounted! We counted some numbers twice.

Alculator11
2020-05-14 16:09:48

What about multiples of 21?

What about multiples of 21?

rrusczyk
2020-05-14 16:09:57

Yeah, the multiples of 21. And what else?

Yeah, the multiples of 21. And what else?

not_kevin_888
2020-05-14 16:10:23

15 and 35

15 and 35

bobthegod78
2020-05-14 16:10:23

35, and 15

35, and 15

pranavaggarwal
2020-05-14 16:10:23

35

35

ZippityA
2020-05-14 16:10:23

and 35

and 35

coldcrazylogic
2020-05-14 16:10:23

15, 35

15, 35

aaronw
2020-05-14 16:10:23

15 and 35

15 and 35

Chesssaga
2020-05-14 16:10:23

Multiples of 15

Multiples of 15

christopherp
2020-05-14 16:10:23

15, 35

15, 35

Doodles250
2020-05-14 16:10:23

and 35

and 35

Dumdum990
2020-05-14 16:10:23

multiples of 15

multiples of 15

tumbleweed
2020-05-14 16:10:23

15 and 35

15 and 35

GrizzyProblemSolver79c
2020-05-14 16:10:23

multiples of 15 and 35

multiples of 15 and 35

rrusczyk
2020-05-14 16:10:32

Yeah, multiples of 15 and 35 are problems, too.

Yeah, multiples of 15 and 35 are problems, too.

rrusczyk
2020-05-14 16:10:55

Let's take a look at what happened with just one of these: $15.$ It has a $3$ and a $5$ in its prime factorization, so it was included twice in our total, once among the multiples of $3$ and once among the multiples of $5.$

Let's take a look at what happened with just one of these: $15.$ It has a $3$ and a $5$ in its prime factorization, so it was included twice in our total, once among the multiples of $3$ and once among the multiples of $5.$

rrusczyk
2020-05-14 16:10:58

The same is true of every multiple of $15.$

The same is true of every multiple of $15.$

rrusczyk
2020-05-14 16:10:59

And of every multiple of $3\cdot 7=21,$ and of every multiple of $5\cdot 7 = 35.$

And of every multiple of $3\cdot 7=21,$ and of every multiple of $5\cdot 7 = 35.$

rrusczyk
2020-05-14 16:11:10

So, what do we do to fix this?

So, what do we do to fix this?

MLiang2018
2020-05-14 16:11:41

subtract

subtract

dineshs
2020-05-14 16:11:41

subtract

subtract

azhang1202
2020-05-14 16:11:41

We subtract them from our total

We subtract them from our total

GoogleNebula
2020-05-14 16:11:41

Subtract

Subtract

obi-wan-the
2020-05-14 16:11:41

SUBTRACT THEM

SUBTRACT THEM

AwesomeLife_Math
2020-05-14 16:11:41

Subtract them from our total.

Subtract them from our total.

NerdyDude
2020-05-14 16:11:41

subtract them

subtract them

RJ5303707
2020-05-14 16:11:41

Subtract these cases

Subtract these cases

Speedstorm
2020-05-14 16:11:41

subtract each of them once

subtract each of them once

robot317
2020-05-14 16:11:41

subtract them

subtract them

doglover07
2020-05-14 16:11:41

subtract those multiples

subtract those multiples

Zoobat
2020-05-14 16:11:41

subtract out those extra ones

subtract out those extra ones

OlympusHero
2020-05-14 16:11:41

Subtract amount of multiples of 15,21 and 35

Subtract amount of multiples of 15,21 and 35

capitanhanbo
2020-05-14 16:11:41

subtract the overlappers

subtract the overlappers

rrusczyk
2020-05-14 16:12:02

We've counted each of these multiples

We've counted each of these multiples

**twice**in our total.
rrusczyk
2020-05-14 16:12:03

So we need to subtract the total number of these multiples

So we need to subtract the total number of these multiples

**once**, to correct for this double counting.
rrusczyk
2020-05-14 16:12:05

How many multiples of $15$ did we double-count?

How many multiples of $15$ did we double-count?

aops-as
2020-05-14 16:12:40

134

134

happyhari
2020-05-14 16:12:40

134

134

Jalenluorion
2020-05-14 16:12:40

134

134

sdattilo2002
2020-05-14 16:12:40

134

134

millburn2006
2020-05-14 16:12:40

134

134

MathBluebird
2020-05-14 16:12:40

134

134

math_piggy
2020-05-14 16:12:40

$134$

$134$

pinkmathcounts
2020-05-14 16:12:40

134

134

sd-pear
2020-05-14 16:12:40

134

134

Elaine_Wang
2020-05-14 16:12:40

134

134

messimagic
2020-05-14 16:12:40

134

134

Math-Marvel
2020-05-14 16:12:40

floor(2019/15)

floor(2019/15)

FireDragon1719
2020-05-14 16:12:40

$floor2019/15$

$floor2019/15$

rrusczyk
2020-05-14 16:12:43

We have $2019 \div 15 = 134\frac{9}{15},$ so there are $134$ multiples of $15$ that we double-counted in our original total.

We have $2019 \div 15 = 134\frac{9}{15},$ so there are $134$ multiples of $15$ that we double-counted in our original total.

rrusczyk
2020-05-14 16:12:44

How about multiples of $21?$

How about multiples of $21?$

melonlord
2020-05-14 16:13:11

96

96

lucky0318
2020-05-14 16:13:11

96

96

scinderella220
2020-05-14 16:13:11

96

96

Math4Life7
2020-05-14 16:13:11

96

96

bryding21
2020-05-14 16:13:11

96

96

AWaltz
2020-05-14 16:13:11

96

96

Quentissential
2020-05-14 16:13:11

96

96

Eth007
2020-05-14 16:13:11

96

96

agentmath
2020-05-14 16:13:11

96

96

KevinW12345
2020-05-14 16:13:11

96

96

DerpyTaterTot
2020-05-14 16:13:11

96

96

PShucks
2020-05-14 16:13:11

96

96

Styxxx
2020-05-14 16:13:11

96

96

ellnoo
2020-05-14 16:13:11

96

96

sarahAops2020
2020-05-14 16:13:11

96

96

kzhang88
2020-05-14 16:13:11

96

96

math12345678
2020-05-14 16:13:11

96

96

PureSwag
2020-05-14 16:13:11

$\lfloor \frac{2019}{21} \rfloor = 96$

$\lfloor \frac{2019}{21} \rfloor = 96$

razormouth
2020-05-14 16:13:11

96

96

badungiface
2020-05-14 16:13:11

96

96

rrusczyk
2020-05-14 16:13:14

We have $2019 \div 21 = 96\frac{3}{21},$ so there are $96$ multiples of $21$ that we double-counted in our original total.

We have $2019 \div 21 = 96\frac{3}{21},$ so there are $96$ multiples of $21$ that we double-counted in our original total.

rrusczyk
2020-05-14 16:13:14

How about multiples of $35?$

How about multiples of $35?$

huela
2020-05-14 16:13:34

57

57

matt1010
2020-05-14 16:13:34

57 for 35

57 for 35

NerdyDude
2020-05-14 16:13:34

57

57

Mandakini
2020-05-14 16:13:34

57

57

bobthefam
2020-05-14 16:13:34

floor(2019/35)=57

floor(2019/35)=57

ZippityA
2020-05-14 16:13:34

57

57

mathlete5451006
2020-05-14 16:13:34

57

57

Eng123
2020-05-14 16:13:34

57

57

bedwinprusik578
2020-05-14 16:13:34

57

57

smartatmath
2020-05-14 16:13:34

57]

57]

Scipow
2020-05-14 16:13:34

57

57

jjw4106
2020-05-14 16:13:34

57

57

srm2005
2020-05-14 16:13:34

57

57

s2010n
2020-05-14 16:13:34

57

57

mathematics10237
2020-05-14 16:13:34

57

57

Knightofaops
2020-05-14 16:13:34

57

57

rrusczyk
2020-05-14 16:13:37

We have $2019 \div 35 = 57\frac{24}{35},$ so there are $57$ multiples of $35$ that we double-counted in our original total.

We have $2019 \div 35 = 57\frac{24}{35},$ so there are $57$ multiples of $35$ that we double-counted in our original total.

rrusczyk
2020-05-14 16:13:39

So, we have to take away $134+96+57 = 287$ from our earlier count, which gives us a new count of $1364 - 287 = 1077.$

So, we have to take away $134+96+57 = 287$ from our earlier count, which gives us a new count of $1364 - 287 = 1077.$

Doudou_Chen
2020-05-14 16:13:54

how about multiples of $105 = 3\cdot5\cdot7$?

how about multiples of $105 = 3\cdot5\cdot7$?

m_goli
2020-05-14 16:13:54

what about 105?

what about 105?

MathWiz20
2020-05-14 16:13:54

what about 105?

what about 105?

NHatAOPS
2020-05-14 16:13:57

what about multiples of 105?

what about multiples of 105?

CESAJ
2020-05-14 16:14:01

but what about multiples of 105?

but what about multiples of 105?

rrusczyk
2020-05-14 16:14:14

What's so important about $105$?

What's so important about $105$?

MathJams
2020-05-14 16:14:37

multiple of 3,5 and 7

multiple of 3,5 and 7

Math5K
2020-05-14 16:14:37

divisible by 3,5,7

divisible by 3,5,7

srim1027
2020-05-14 16:14:37

it is the lcm of 3 5 7

it is the lcm of 3 5 7

MyNameIsJeffZF
2020-05-14 16:14:37

3*5*7

3*5*7

mathcounts_genius
2020-05-14 16:14:37

3*5*7

3*5*7

mmjguitar
2020-05-14 16:14:37

its 5*3*7

its 5*3*7

yayatheduck
2020-05-14 16:14:37

it's 3*5*7

it's 3*5*7

anshultheking1
2020-05-14 16:14:37

3x5x7

3x5x7

Shmileyface
2020-05-14 16:14:37

multiple of 3, 5, and 7

multiple of 3, 5, and 7

olive0827
2020-05-14 16:14:37

3*5*7

3*5*7

enkitty
2020-05-14 16:14:37

It's a multiple of all three.

It's a multiple of all three.

huangtx2018
2020-05-14 16:14:39

it's the LCM of 3, 5, and 7

it's the LCM of 3, 5, and 7

DonutHole
2020-05-14 16:14:39

It is a multiple of 3, 5, and 7

It is a multiple of 3, 5, and 7

rrusczyk
2020-05-14 16:14:44

It is a multiple of all of $3, 5,$ and $7.$

It is a multiple of all of $3, 5,$ and $7.$

rrusczyk
2020-05-14 16:14:53

Hmmmm... How many times have we counted it so far?

Hmmmm... How many times have we counted it so far?

huela
2020-05-14 16:15:18

0

0

scoutskylar
2020-05-14 16:15:18

0

0

RJ5303707
2020-05-14 16:15:18

None

None

zhuqingzhang
2020-05-14 16:15:18

0

0

Math5K
2020-05-14 16:15:18

0

0

rayfish
2020-05-14 16:15:18

0 times!!!

0 times!!!

Pokemon2
2020-05-14 16:15:18

0 times

0 times

missionsqhc
2020-05-14 16:15:18

0

0

rrusczyk
2020-05-14 16:15:38

None? But I'm pretty sure we counted it with the multiples of 3. And the multiples of 5. And the multiples of 7.

None? But I'm pretty sure we counted it with the multiples of 3. And the multiples of 5. And the multiples of 7.

scoutskylar
2020-05-14 16:16:04

We counted them three times, but un-counted them three times.

We counted them three times, but un-counted them three times.

arirah9
2020-05-14 16:16:04

3 times, and then we subtracted it 3 times, so 0

3 times, and then we subtracted it 3 times, so 0

DottedCaculator
2020-05-14 16:16:04

but then we subtracted them

but then we subtracted them

odinthunder
2020-05-14 16:16:04

But then we subtracted 3 times

But then we subtracted 3 times

green_alligator
2020-05-14 16:16:04

but we subtracted it 3 times as well

but we subtracted it 3 times as well

lilyba
2020-05-14 16:16:06

we then subtracted it three times

we then subtracted it three times

user0003
2020-05-14 16:16:08

subtracted three times

subtracted three times

rrusczyk
2020-05-14 16:16:38

Oh, yeah. We took it back away with the multiples of 15. And the multiples of 21. And the multiples of 35.

Oh, yeah. We took it back away with the multiples of 15. And the multiples of 21. And the multiples of 35.

SaltyCracker
2020-05-14 16:16:48

it overlaps with the other overlappers

it overlaps with the other overlappers

rrusczyk
2020-05-14 16:16:56

I could not possibly have said it better myself.

I could not possibly have said it better myself.

rrusczyk
2020-05-14 16:17:05

It got counted three times in our original count (once for each prime), and then subtracted three times in our correction (once for each product of two primes).

It got counted three times in our original count (once for each prime), and then subtracted three times in our correction (once for each product of two primes).

rrusczyk
2020-05-14 16:17:07

That means that in our total of $1077,$ the number $105$ (and all multiples of $105$) aren't counted at all!

That means that in our total of $1077,$ the number $105$ (and all multiples of $105$) aren't counted at all!

Mistyketchum28
2020-05-14 16:17:20

We have to add it back once!

We have to add it back once!

DonutHole
2020-05-14 16:17:20

So we have to add it back

So we have to add it back

anishcool11
2020-05-14 16:17:24

add them back

add them back

nikenissan
2020-05-14 16:17:26

We need to add back that number of 105 multiples

We need to add back that number of 105 multiples

mark888
2020-05-14 16:17:34

oh no, gotta add it back

oh no, gotta add it back

Chesssaga
2020-05-14 16:17:34

We need to add back

We need to add back

bluemathcounts
2020-05-14 16:17:36

add back

add back

poopypurplebob
2020-05-14 16:17:38

add it back

add it back

rrusczyk
2020-05-14 16:17:41

We have to add them back in.

We have to add them back in.

fasterthanlight
2020-05-14 16:17:51

There are 19 of them

There are 19 of them

NHatAOPS
2020-05-14 16:17:51

there's 19 multiples

there's 19 multiples

MLTC
2020-05-14 16:17:51

So add 19 to our answer

So add 19 to our answer

jupiter314
2020-05-14 16:17:51

add back 19

add back 19

CHIPPER33
2020-05-14 16:17:51

add 19!

add 19!

MathMaster029
2020-05-14 16:17:51

add 19

add 19

Maths4J
2020-05-14 16:17:51

There are 19 of them

There are 19 of them

enkitty
2020-05-14 16:17:51

There are 19 multiples of 105, so add it to 1077

There are 19 multiples of 105, so add it to 1077

Asterlan
2020-05-14 16:17:53

19 multiples of it to add back!

19 multiples of it to add back!

rrusczyk
2020-05-14 16:17:56

Since $2019 \div 105 = 19\frac{24}{105},$ we have $19$ multiples of $105$ less than $2019.$ (It's probably easier to just notice that $20 \cdot 105 = 2100$ is too big, and $19 \cdot 105 = 2100 - 105 = 1995$ is small enough.)

Since $2019 \div 105 = 19\frac{24}{105},$ we have $19$ multiples of $105$ less than $2019.$ (It's probably easier to just notice that $20 \cdot 105 = 2100$ is too big, and $19 \cdot 105 = 2100 - 105 = 1995$ is small enough.)

rrusczyk
2020-05-14 16:17:59

We add these back in to our running total and we have $1077+19 = 1096$ total numbers with at least one odd single-digit prime factor.

We add these back in to our running total and we have $1077+19 = 1096$ total numbers with at least one odd single-digit prime factor.

rrusczyk
2020-05-14 16:18:12

Phew, so do we write down 1096 and move on now?

Phew, so do we write down 1096 and move on now?

HappySpring
2020-05-14 16:18:47

we have to subtract it from 2019

we have to subtract it from 2019

mathboykl2018
2020-05-14 16:18:47

subtract that from 2019

subtract that from 2019

AwesomeLife_Math
2020-05-14 16:18:47

Subtract that from 2019 to get the answer.

Subtract that from 2019 to get the answer.

ARSM2019
2020-05-14 16:18:47

2019-1096

2019-1096

Jomo
2020-05-14 16:18:47

wait! we are complementary counting!!!

wait! we are complementary counting!!!

bobthegod78
2020-05-14 16:18:47

2019-1096=923!!

2019-1096=923!!

NHatAOPS
2020-05-14 16:18:47

subtract that from 2019 to get the answer

subtract that from 2019 to get the answer

joseph2718
2020-05-14 16:18:47

2019 - 1096 !!

2019 - 1096 !!

IMadeYouReadThis
2020-05-14 16:18:47

Now do $2019-1096$

Now do $2019-1096$

anc3
2020-05-14 16:18:47

Noooo subtract from total of 2019 because we used complementary counting

Noooo subtract from total of 2019 because we used complementary counting

palindrome868
2020-05-14 16:18:47

2019 - 1096 = 923!!!

2019 - 1096 = 923!!!

RedFireTruck
2020-05-14 16:18:48

2019-1096 = 923

2019-1096 = 923

rrusczyk
2020-05-14 16:18:56

Oh yeah! I'm glad y'all are on my team!

Oh yeah! I'm glad y'all are on my team!

rrusczyk
2020-05-14 16:19:02

We've now accurately counted each multiple of $3, 5,$ or $7$ exactly once.

We've now accurately counted each multiple of $3, 5,$ or $7$ exactly once.

rrusczyk
2020-05-14 16:19:04

But remember these are the numbers we

But remember these are the numbers we

*don't*want!
rrusczyk
2020-05-14 16:19:05

We have to subtract this count from $2019$ to get our final answer.

We have to subtract this count from $2019$ to get our final answer.

PShucks
2020-05-14 16:19:15

the answer is 923

the answer is 923

akpi
2020-05-14 16:19:24

2019-1096=923

2019-1096=923

lordoftherings4926
2020-05-14 16:19:24

We write down 923

We write down 923

Speedstorm
2020-05-14 16:19:24

2019-1096=923

2019-1096=923

hunthunthunt
2020-05-14 16:19:24

$923$

$923$

dhaops
2020-05-14 16:19:29

923.

923.

jonahc426
2020-05-14 16:19:29

923

923

Raiders26
2020-05-14 16:19:29

923

923

sixoneeight
2020-05-14 16:19:30

So, the answer is 923

So, the answer is 923

Dumdum990
2020-05-14 16:19:30

923 possible outcomes

923 possible outcomes

robot317
2020-05-14 16:19:33

the answer is 923

the answer is 923

rrusczyk
2020-05-14 16:19:35

Therefore, $2019-1096 = \boxed{923}$ of the first $2019$ positive integers have no odd single-digit prime factors.

Therefore, $2019-1096 = \boxed{923}$ of the first $2019$ positive integers have no odd single-digit prime factors.

ZOP123
2020-05-14 16:19:44

YAY

YAY

Math4Life7
2020-05-14 16:19:44

yayyyyy

yayyyyy

rrusczyk
2020-05-14 16:19:53

All right -- DPatrick's turn!

All right -- DPatrick's turn!

DPatrick
2020-05-14 16:19:59

Thanks -- hi again!

Thanks -- hi again!

DPatrick
2020-05-14 16:20:07

**Sprint #28:**What is the value of $\sqrt{11{,}111{,}111\times 100{,}000{,}011+4}?$
DPatrick
2020-05-14 16:20:21

Remember: no calculators!

Remember: no calculators!

nathanqiu
2020-05-14 16:20:47

that looks terrifying!

that looks terrifying!

kaciecheng
2020-05-14 16:20:50

ugh big headache . -.

ugh big headache . -.

DPatrick
2020-05-14 16:20:59

Yikes. I don't want to multiply out those big numbers, add $4,$ and try to take a square root.

Yikes. I don't want to multiply out those big numbers, add $4,$ and try to take a square root.

DPatrick
2020-05-14 16:21:09

Any better ideas?

Any better ideas?

sixoneeight
2020-05-14 16:21:31

Try to find a pattern!

Try to find a pattern!

menlo
2020-05-14 16:21:42

100000011=100000000+11

100000011=100000000+11

DPatrick
2020-05-14 16:22:01

I like this thought. Are there easier ways we can write those big numbers?

I like this thought. Are there easier ways we can write those big numbers?

Makemakika1
2020-05-14 16:22:15

11111111 is about (1/9)×10^8, the other number is about 10^8.

11111111 is about (1/9)×10^8, the other number is about 10^8.

iiRishabii
2020-05-14 16:22:31

Split 100000011 into (100000000 + 11)

Split 100000011 into (100000000 + 11)

Qiu10
2020-05-14 16:22:36

100000011 is 100000000+11

100000011 is 100000000+11

duansaops
2020-05-14 16:22:36

11111111 = (10^8-1)/9

11111111 = (10^8-1)/9

DPatrick
2020-05-14 16:22:45

Aha!

Aha!

DPatrick
2020-05-14 16:22:57

$11{,}111{,}111$ is one-ninth of $99{,}999{,}999,$ which is one less than $100{,}000{,}000 = 10^8.$

$11{,}111{,}111$ is one-ninth of $99{,}999{,}999,$ which is one less than $100{,}000{,}000 = 10^8.$

DPatrick
2020-05-14 16:23:16

So we can write $11{,}111{,}111 = \dfrac{10^8-1}{9}.$

So we can write $11{,}111{,}111 = \dfrac{10^8-1}{9}.$

DPatrick
2020-05-14 16:23:25

That seems like it'll be a lot easier to work with.

That seems like it'll be a lot easier to work with.

Jomo
2020-05-14 16:23:40

and $10^8+11$

and $10^8+11$

sixoneeight
2020-05-14 16:23:40

so its 10^8+11

so its 10^8+11

MR_67
2020-05-14 16:23:40

100000011 = 10^8+11, 11111111=(10^8-1)/9

100000011 = 10^8+11, 11111111=(10^8-1)/9

DPatrick
2020-05-14 16:23:51

Good idea. Especially since we have $10^8$ in our first number, I think $10^8 + 11$ is a lot easier to work with for the second number.

Good idea. Especially since we have $10^8$ in our first number, I think $10^8 + 11$ is a lot easier to work with for the second number.

DPatrick
2020-05-14 16:24:03

So now our expression is $$\sqrt{\dfrac{10^8-1}{9} \times (10^8+11) + 4}.$$

So now our expression is $$\sqrt{\dfrac{10^8-1}{9} \times (10^8+11) + 4}.$$

DPatrick
2020-05-14 16:24:12

This seems much simpler to work with!

This seems much simpler to work with!

DPatrick
2020-05-14 16:24:19

What now?

What now?

Alexm12
2020-05-14 16:24:47

9 is a square!!

9 is a square!!

ajax31
2020-05-14 16:24:47

9 is 3²

9 is 3²

Scipow
2020-05-14 16:24:47

factor out the 1/9!

factor out the 1/9!

Ninja-Girl
2020-05-14 16:25:00

$9$ = $3^2$

$9$ = $3^2$

enkitty
2020-05-14 16:25:03

Could we multiply by 9 so we don't have to deal with fractions

Could we multiply by 9 so we don't have to deal with fractions

DPatrick
2020-05-14 16:25:19

Well, that $9$ is a tantalizing perfect square just sitting there.

Well, that $9$ is a tantalizing perfect square just sitting there.

DPatrick
2020-05-14 16:25:36

But it's hard to work with, seeing as it's sitting there by itself.

But it's hard to work with, seeing as it's sitting there by itself.

scoutskylar
2020-05-14 16:25:45

That's equal to $\frac{1}{3}\sqrt{\left(10^{8}-1\right)\cdot\left(10^{8}+11\right)+36}$

That's equal to $\frac{1}{3}\sqrt{\left(10^{8}-1\right)\cdot\left(10^{8}+11\right)+36}$

DottedCaculator
2020-05-14 16:25:57

4=36/9

4=36/9

Alculator11
2020-05-14 16:25:59

Multiply $4$ by $9/9$

Multiply $4$ by $9/9$

Raiders26
2020-05-14 16:26:10

make denominator 9

make denominator 9

DPatrick
2020-05-14 16:26:16

Good idea.

Good idea.

DPatrick
2020-05-14 16:26:26

Let's make it a single fraction!

Let's make it a single fraction!

DPatrick
2020-05-14 16:26:31

We get $\sqrt{\dfrac{(10^8-1)(10^8+11)+36}{9}}.$

We get $\sqrt{\dfrac{(10^8-1)(10^8+11)+36}{9}}.$

DPatrick
2020-05-14 16:26:47

And now we can pull the $9$ outside if we want.

And now we can pull the $9$ outside if we want.

DPatrick
2020-05-14 16:27:08

So we have $\dfrac13\sqrt{(10^8-1)(10^8+11)+36}.$

So we have $\dfrac13\sqrt{(10^8-1)(10^8+11)+36}.$

DPatrick
2020-05-14 16:27:19

Now what?

Now what?

asbodke
2020-05-14 16:27:28

Multiply it out

Multiply it out

MathJams
2020-05-14 16:27:28

expand $(10^8-1)(10^8+11)+36$!

expand $(10^8-1)(10^8+11)+36$!

ThatRichDeng
2020-05-14 16:27:36

expand it

expand it

dolphin7
2020-05-14 16:27:36

expand

expand

doglover07
2020-05-14 16:27:36

expand

expand

user0003
2020-05-14 16:27:36

expand

expand

agentmath
2020-05-14 16:27:36

expand

expand

melonlord
2020-05-14 16:27:36

expand

expand

KingRavi
2020-05-14 16:27:36

expand

expand

DPatrick
2020-05-14 16:27:48

Sure, we can multiply the two binomials together.

Sure, we can multiply the two binomials together.

DPatrick
2020-05-14 16:27:57

What do we get under the square root when we do that?

What do we get under the square root when we do that?

Speedstorm
2020-05-14 16:28:44

The root simplifies to $10^{16}+10^9+25$

The root simplifies to $10^{16}+10^9+25$

AbhiWwis
2020-05-14 16:28:44

10^16 + 10(10^8) + 25

10^16 + 10(10^8) + 25

peace09
2020-05-14 16:28:44

$10^16+10^9+5^2$

$10^16+10^9+5^2$

sigma_notation
2020-05-14 16:28:44

multiply it out to (10^16+10^9+25)

multiply it out to (10^16+10^9+25)

palindrome868
2020-05-14 16:28:44

10^16 + 10^9 + 25

10^16 + 10^9 + 25

DPatrick
2020-05-14 16:29:01

We get $\dfrac13\sqrt{(10^8)^2 + 10(10^8) + 25}.$

We get $\dfrac13\sqrt{(10^8)^2 + 10(10^8) + 25}.$

MTHJJS
2020-05-14 16:29:33

the square root is = $(10^8 + 5)$

the square root is = $(10^8 + 5)$

AbhiWwis
2020-05-14 16:29:33

(10^8 + 5)^2

(10^8 + 5)^2

Speedstorm
2020-05-14 16:29:33

$(10^8+5)^2$

$(10^8+5)^2$

duansaops
2020-05-14 16:29:33

$(10^8+5)^2$

$(10^8+5)^2$

Maths4J
2020-05-14 16:29:33

(10^8+5)^2

(10^8+5)^2

sdattilo2002
2020-05-14 16:29:33

nice!

nice!

srim1027
2020-05-14 16:29:33

square of a binomial

square of a binomial

Quentissential
2020-05-14 16:29:33

(10^8 +5)^2

(10^8 +5)^2

Dalar25
2020-05-14 16:29:33

perfect square!

perfect square!

nascar48
2020-05-14 16:29:33

perfect square

perfect square

DPatrick
2020-05-14 16:29:39

Hey, that looks like a perfect square!

Hey, that looks like a perfect square!

DPatrick
2020-05-14 16:29:55

Indeed, it's $\dfrac13\sqrt{\left(10^8 + 5\right)^2}.$

Indeed, it's $\dfrac13\sqrt{\left(10^8 + 5\right)^2}.$

Imishkabob
2020-05-14 16:30:02

And then we take the perfect squares out

And then we take the perfect squares out

Math5K
2020-05-14 16:30:06

it's 1/3(10^8+5)

it's 1/3(10^8+5)

sixoneeight
2020-05-14 16:30:14

Wow! so we get (10^8+5)/3

Wow! so we get (10^8+5)/3

vtlev
2020-05-14 16:30:16

(10^8+5)/3

(10^8+5)/3

DPatrick
2020-05-14 16:30:20

So our expression is just $\dfrac{10^8 + 5}{3}.$

So our expression is just $\dfrac{10^8 + 5}{3}.$

DPatrick
2020-05-14 16:30:28

And how do we finish up?

And how do we finish up?

superagh
2020-05-14 16:30:42

and then we divide

and then we divide

DottedCaculator
2020-05-14 16:30:42

=10000005/3=33333335

=10000005/3=33333335

MR_67
2020-05-14 16:30:42

100000005/3 = 33333335

100000005/3 = 33333335

athik789
2020-05-14 16:30:42

100000005/3

100000005/3

Mistyketchum28
2020-05-14 16:30:45

100000005/3

100000005/3

Ninja11
2020-05-14 16:30:45

which is 100000005/3

which is 100000005/3

GrizzyProblemSolver79c
2020-05-14 16:30:54

$100000005/3 = 33333335$

$100000005/3 = 33333335$

IMadeYouReadThis
2020-05-14 16:30:54

$\frac{100000005}{3}$

$\frac{100000005}{3}$

Lux1
2020-05-14 16:30:54

and then that comes out to 33,333,335

and then that comes out to 33,333,335

DPatrick
2020-05-14 16:31:00

We could certainly compute $100{,}000{,}005 \div 3,$ but let's use the same trick we used in the beginning.

We could certainly compute $100{,}000{,}005 \div 3,$ but let's use the same trick we used in the beginning.

DPatrick
2020-05-14 16:31:07

We know that $\dfrac{10^8-1}{3} = 33{,}333{,}333.$ How does that help?

We know that $\dfrac{10^8-1}{3} = 33{,}333{,}333.$ How does that help?

b20081
2020-05-14 16:31:19

note that 9999999=333333*3

note that 9999999=333333*3

BakedPotato66
2020-05-14 16:31:23

$10^8-1$ is divisible by 3

$10^8-1$ is divisible by 3

b20081
2020-05-14 16:31:29

and then 9999999+6=10^8+5

and then 9999999+6=10^8+5

GrizzyProblemSolver79c
2020-05-14 16:31:32

it might be easier to do $(99999999 + 6)/3$

it might be easier to do $(99999999 + 6)/3$

PureSwag
2020-05-14 16:31:35

We can note that $99,999,999 \div 3 = 33,333,333$

We can note that $99,999,999 \div 3 = 33,333,333$

MyNameIsJeffZF
2020-05-14 16:31:40

add 6

add 6

Wontoflonto
2020-05-14 16:31:40

add 6/3 = 2

add 6/3 = 2

huela
2020-05-14 16:31:40

+6/3

+6/3

Puffer13
2020-05-14 16:31:40

just add 6/3

just add 6/3

DPatrick
2020-05-14 16:31:43

Exactly!

Exactly!

DPatrick
2020-05-14 16:31:51

We can write $$\dfrac{10^8+5}{3} = \dfrac{10^8-1+6}{3} = \dfrac{10^8-1}{3} + 2.$$

We can write $$\dfrac{10^8+5}{3} = \dfrac{10^8-1+6}{3} = \dfrac{10^8-1}{3} + 2.$$

aaravdodhia
2020-05-14 16:32:10

Then $\frac{10^8 + 5}{3} = \frac{10^8 - 1}{3} + \frac{6}{3} = 33,333,333 + 2$.

Then $\frac{10^8 + 5}{3} = \frac{10^8 - 1}{3} + \frac{6}{3} = 33,333,333 + 2$.

C0atimundi
2020-05-14 16:32:10

By adding 6 on the top we add 2 to 33,333,333

By adding 6 on the top we add 2 to 33,333,333

Ninja11
2020-05-14 16:32:10

You can add 2 to 33333333

You can add 2 to 33333333

dolphin7
2020-05-14 16:32:15

$33,333,333+2=33,333,335$

$33,333,333+2=33,333,335$

CosmoMonkhouse
2020-05-14 16:32:20

All we have to do is add 2 to get 33333335

All we have to do is add 2 to get 33333335

bobjoebilly
2020-05-14 16:32:20

33, 333, 335

33, 333, 335

DPatrick
2020-05-14 16:32:24

And then this is just $33{,}333{,}333 + 2 = \boxed{33{,}333{,}335}.$

And then this is just $33{,}333{,}333 + 2 = \boxed{33{,}333{,}335}.$

DPatrick
2020-05-14 16:32:45

The moral of the story: big numbers are icky. Try to make them less icky.

The moral of the story: big numbers are icky. Try to make them less icky.

bestzack66
2020-05-14 16:32:55

YAY we solved another problem!

YAY we solved another problem!

Speedstorm
2020-05-14 16:32:55

that was a nice solution

that was a nice solution

ssr_07
2020-05-14 16:32:55

That was a good one.

That was a good one.

MrAnishNagariya
2020-05-14 16:32:59

that looked so scary, not it seems so easy

that looked so scary, not it seems so easy

DPatrick
2020-05-14 16:33:08

A lot of problems look scarier than they really are!

A lot of problems look scarier than they really are!

DPatrick
2020-05-14 16:33:23

I feel like the next problem was written especially for me!

I feel like the next problem was written especially for me!

DPatrick
2020-05-14 16:33:29

**Sprint #29:**David throws a dart at a triangular dartboard whose side lengths are $5, 5$ and $6,$ and the dart lands in a random location on the dartboard. What is the probability that the sum of the squares of the $3$ distances from the dart's location to the corners of the dartboard is less than $30?$ Express your answer as a common fraction in terms of $\pi.$
Eng123
2020-05-14 16:34:02

Geometric probability?

Geometric probability?

CHIPPER33
2020-05-14 16:34:02

Geometric Probability!!

Geometric Probability!!

Pleaseletmewin
2020-05-14 16:34:02

Is Harvey here for geometric probability?

Is Harvey here for geometric probability?

Equinox8
2020-05-14 16:34:02

Geometric probability

Geometric probability

DPatrick
2020-05-14 16:34:19

Harvey's on break with rrusczyk, but fortunately I think (I hope!) I can handle geometric probability.

Harvey's on break with rrusczyk, but fortunately I think (I hope!) I can handle geometric probability.

DPatrick
2020-05-14 16:34:29

Why does "geometric probability" feel like it'll be the right thing to try here?

Why does "geometric probability" feel like it'll be the right thing to try here?

walrus987
2020-05-14 16:35:08

because it's geometric and it's probability?

because it's geometric and it's probability?

eibc
2020-05-14 16:35:08

because there are infintely many possibilities

because there are infintely many possibilities

FALCONSRULE1
2020-05-14 16:35:08

it is geometry and probabality

it is geometry and probabality

missionsqhc
2020-05-14 16:35:08

We have infinitely many points.

We have infinitely many points.

cubi
2020-05-14 16:35:08

can't count all the possible positions

can't count all the possible positions

DPatrick
2020-05-14 16:35:26

Right. There are infinitely many places the dart could land: any point on the dartboard is fair game!

Right. There are infinitely many places the dart could land: any point on the dartboard is fair game!

DPatrick
2020-05-14 16:35:33

So we can't possibly hope to count them.

So we can't possibly hope to count them.

rayfish
2020-05-14 16:35:57

$\text{probability}=\frac{\text{area of successful region}}{\text{area of triangle}}$

$\text{probability}=\frac{\text{area of successful region}}{\text{area of triangle}}$

aaravdodhia
2020-05-14 16:36:10

Probability = $\frac{\text{Successful Area}}{\text{Possible Area}}$.

Probability = $\frac{\text{Successful Area}}{\text{Possible Area}}$.

subr3587835
2020-05-14 16:36:21

Area of success/total area

Area of success/total area

DPatrick
2020-05-14 16:36:23

Right! To compute the probability, we'll want to use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$

Right! To compute the probability, we'll want to use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$

DPatrick
2020-05-14 16:36:35

So we need to figure out what those regions are.

So we need to figure out what those regions are.

mathuser18
2020-05-14 16:36:47

Can we draw a diagram?

Can we draw a diagram?

srim1027
2020-05-14 16:36:47

draw a diagram?

draw a diagram?

ZippityA
2020-05-14 16:36:47

We should draw a diagram first.

We should draw a diagram first.

MathHayden
2020-05-14 16:36:47

first we draw the dartboard

first we draw the dartboard

Raiders26
2020-05-14 16:36:47

draw an area of solution on the dartboard

draw an area of solution on the dartboard

DPatrick
2020-05-14 16:37:05

Yep, we're definitely going to need to draw a picture!

Yep, we're definitely going to need to draw a picture!

Pokemon2
2020-05-14 16:37:13

could we coordinate bash?

could we coordinate bash?

CHIPPER33
2020-05-14 16:37:27

Create a coordinate system with the triangle, There are infinite points to calculate

Create a coordinate system with the triangle, There are infinite points to calculate

Puffer13
2020-05-14 16:37:37

set the triangle on the coordinate plane

set the triangle on the coordinate plane

mjl09
2020-05-14 16:37:48

coordinates system

coordinates system

FALCONSRULE1
2020-05-14 16:37:48

coordinate plane

coordinate plane

DPatrick
2020-05-14 16:37:58

Good idea. We're going to need to compute some distances, so setting this up on the coordinate plane may be helpful.

Good idea. We're going to need to compute some distances, so setting this up on the coordinate plane may be helpful.

DPatrick
2020-05-14 16:38:21

Any suggestions how to set it up?

Any suggestions how to set it up?

Imishkabob
2020-05-14 16:38:46

0,0 could be one point

0,0 could be one point

fasterthanlight
2020-05-14 16:38:46

Set one corner to be (0,0)

Set one corner to be (0,0)

Jomo
2020-05-14 16:38:46

set center of base at origin

set center of base at origin

Asterlan
2020-05-14 16:38:46

One corner 0,0

One corner 0,0

KRISHIV
2020-05-14 16:38:46

set one point on (0,0) and work from there

set one point on (0,0) and work from there

DPatrick
2020-05-14 16:38:56

The origin is a really convenient point of the coordinate plane!

The origin is a really convenient point of the coordinate plane!

DPatrick
2020-05-14 16:39:06

So let's put the base of the dartboard along the $x$-axis, so that one corner is at $(0,0)$ and one corner is at $(6,0)$:

So let's put the base of the dartboard along the $x$-axis, so that one corner is at $(0,0)$ and one corner is at $(6,0)$:

DPatrick
2020-05-14 16:39:12

DPatrick
2020-05-14 16:39:16

What point is the top of the dartboard?

What point is the top of the dartboard?

AlexWangMath
2020-05-14 16:39:38

3-4-5 triangles

3-4-5 triangles

Math5K
2020-05-14 16:39:38

triangle vertices: (0,0),(3,4),(6,0)

triangle vertices: (0,0),(3,4),(6,0)

kdraganov
2020-05-14 16:39:38

(0,0) -- (3,4) -- (0,6)

(0,0) -- (3,4) -- (0,6)

Pleaseletmewin
2020-05-14 16:39:38

Wannabe 3-4-5 triangle

Wannabe 3-4-5 triangle

IMadeYouReadThis
2020-05-14 16:39:38

Set the points at $(0,0)$, $(6,0)$, and $(3,4)$.

Set the points at $(0,0)$, $(6,0)$, and $(3,4)$.

awesomeming327.
2020-05-14 16:39:38

3,4

3,4

Feifei01
2020-05-14 16:39:38

(3, 4)

(3, 4)

kdraganov
2020-05-14 16:39:38

$(3,4)$

$(3,4)$

asbodke
2020-05-14 16:39:38

(3,4)

(3,4)

AMC_Kid
2020-05-14 16:39:38

(3,4)

(3,4)

Math5K
2020-05-14 16:39:38

(3,4)

(3,4)

cai40
2020-05-14 16:39:38

(3,4)

(3,4)

bobamilkcha
2020-05-14 16:39:38

(3,4)

(3,4)

yayatheduck
2020-05-14 16:39:38

3,4

3,4

punkinpiday
2020-05-14 16:39:38

(3,4)

(3,4)

sixoneeight
2020-05-14 16:39:38

3,4

3,4

GrizzyProblemSolver79c
2020-05-14 16:39:38

(3, 4)

(3, 4)

DPatrick
2020-05-14 16:39:50

Yes, 3-4-5 triangles FTW! Drawing the altitude of the dartboard to the side of length $6$ divides the board into two $3$-$4$-$5$ right triangles, so the top point is the point $(3,4)$:

Yes, 3-4-5 triangles FTW! Drawing the altitude of the dartboard to the side of length $6$ divides the board into two $3$-$4$-$5$ right triangles, so the top point is the point $(3,4)$:

DPatrick
2020-05-14 16:40:01

DPatrick
2020-05-14 16:40:33

So remember, what we're trying to do is use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$

So remember, what we're trying to do is use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$

DPatrick
2020-05-14 16:40:43

Once of these is pretty easy.

Once of these is pretty easy.

Mrs. H math teacher
2020-05-14 16:41:05

Area of the triangle is 12.

Area of the triangle is 12.

happyhari
2020-05-14 16:41:05

the denominator is 12

the denominator is 12

ssr_07
2020-05-14 16:41:05

So height is 4 and base is 6. The area is 12 units squared.

So height is 4 and base is 6. The area is 12 units squared.

Chesssaga
2020-05-14 16:41:05

Area of possible region = 12

Area of possible region = 12

cat_maniac_8
2020-05-14 16:41:05

12

12

rneelam
2020-05-14 16:41:05

its 12

its 12

MathBluebird
2020-05-14 16:41:05

area of triangle = 12

area of triangle = 12

Wontoflonto
2020-05-14 16:41:05

area of the possible region, 12

area of the possible region, 12

cj13609517288
2020-05-14 16:41:05

possible region is just 12

possible region is just 12

MathIsFun286
2020-05-14 16:41:05

area of possible region is $12$

area of possible region is $12$

song2sons
2020-05-14 16:41:05

area of possible region is 12

area of possible region is 12

millburn2006
2020-05-14 16:41:05

area of possible region = 12

area of possible region = 12

DPatrick
2020-05-14 16:41:29

The "possible region" is the entire dartboard, because that's where the dart could land.

The "possible region" is the entire dartboard, because that's where the dart could land.

DPatrick
2020-05-14 16:41:38

So the denominator of our probability is just the area of the triangle.

So the denominator of our probability is just the area of the triangle.

DPatrick
2020-05-14 16:41:46

Since the base has length $6$ and the height is $4,$ its area is $\frac12(6)(4) = 12.$

Since the base has length $6$ and the height is $4,$ its area is $\frac12(6)(4) = 12.$

DPatrick
2020-05-14 16:41:56

Now for the harder part. How do we determine the area of the "successful" region?

Now for the harder part. How do we determine the area of the "successful" region?

winterrain01
2020-05-14 16:42:30

Use the distance formula?

Use the distance formula?

manokid
2020-05-14 16:42:30

Distance formula

Distance formula

ant08
2020-05-14 16:42:30

distance formula

distance formula

vsamc
2020-05-14 16:42:30

Pick an arbitrary point $(a,b)$ inside the triangle

Pick an arbitrary point $(a,b)$ inside the triangle

CHIPPER33
2020-05-14 16:42:30

Call the random point (x,y) and add up the distance squared to vertexes.

Call the random point (x,y) and add up the distance squared to vertexes.

dudavid
2020-05-14 16:42:30

Distance formula

Distance formula

Meyzeek_Saveer
2020-05-14 16:42:30

distance formula

distance formula

DPatrick
2020-05-14 16:42:48

Right! Suppose the dart lands at some point $(x,y).$

Right! Suppose the dart lands at some point $(x,y).$

DPatrick
2020-05-14 16:42:59

We can write an expression for the sum of the squares of the distances from the dart to the corners of the board, using the distance formula.

We can write an expression for the sum of the squares of the distances from the dart to the corners of the board, using the distance formula.

DPatrick
2020-05-14 16:43:04

What's the square of the distance from $(x,y)$ to $(0,0)?$

What's the square of the distance from $(x,y)$ to $(0,0)?$

Maths4J
2020-05-14 16:43:24

x^2+y^2

x^2+y^2

Imayormaynotknowcalculus
2020-05-14 16:43:24

$x^2+y^2$

$x^2+y^2$

TheEpicCarrot7
2020-05-14 16:43:24

x^2+y^2

x^2+y^2

Alculator11
2020-05-14 16:43:24

$x^2+y^2$

$x^2+y^2$

Speedstorm
2020-05-14 16:43:24

$x^2+y^2$

$x^2+y^2$

HappySpring
2020-05-14 16:43:24

sqrt (x^2 + y^2)

sqrt (x^2 + y^2)

Jerry_Guo
2020-05-14 16:43:24

x^2+y^2

x^2+y^2

scoutskylar
2020-05-14 16:43:24

$x^2+y^2$

$x^2+y^2$

yukrant1
2020-05-14 16:43:24

$x^2+y^2$

$x^2+y^2$

DPatrick
2020-05-14 16:43:31

It's $x^2 + y^2.$

It's $x^2 + y^2.$

DPatrick
2020-05-14 16:43:34

What's the square of the distance from $(x,y)$ to $(6,0)?$

What's the square of the distance from $(x,y)$ to $(6,0)?$

doglover07
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

Leonard_my_dude
2020-05-14 16:43:54

(x-6)^2 + y^2

(x-6)^2 + y^2

melonlord
2020-05-14 16:43:54

(6-x)^2 +y^2

(6-x)^2 +y^2

Gumball
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

nathanqiu
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

mathcounts_genius
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

Shmileyface
2020-05-14 16:43:54

(x-6)^2 + y^2

(x-6)^2 + y^2

SharonW
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

Puffer13
2020-05-14 16:43:54

(x-6)^2+y^2

(x-6)^2+y^2

DPatrick
2020-05-14 16:44:00

It's $(x-6)^2 + y^2.$

It's $(x-6)^2 + y^2.$

DPatrick
2020-05-14 16:44:04

What's the square of the distance from $(x,y)$ to $(3,4)?$

What's the square of the distance from $(x,y)$ to $(3,4)?$

huela
2020-05-14 16:44:24

(x-3)^2+(y-4)^2

(x-3)^2+(y-4)^2

lrjr24
2020-05-14 16:44:24

$(x-3)^2+(y-4)^2$

$(x-3)^2+(y-4)^2$

menlo
2020-05-14 16:44:24

(x-3)^2 + (y-4)^2

(x-3)^2 + (y-4)^2

Kruxe
2020-05-14 16:44:24

(x - 3)^2 + (y - 4)^2

(x - 3)^2 + (y - 4)^2

C0atimundi
2020-05-14 16:44:24

(x-3)^2 * (y-4)^2

(x-3)^2 * (y-4)^2

walrus987
2020-05-14 16:44:24

(x-3)^2 + (y-4)^2

(x-3)^2 + (y-4)^2

ellnoo
2020-05-14 16:44:24

(x-3)^2+(y-4)^2

(x-3)^2+(y-4)^2

srim1027
2020-05-14 16:44:24

(x-3)^2 + (y-4)^2

(x-3)^2 + (y-4)^2

NerdyDude
2020-05-14 16:44:24

(x-3)^2+(y-4)^2

(x-3)^2+(y-4)^2

jai123
2020-05-14 16:44:24

(x-3)^2+(y-4)^2

(x-3)^2+(y-4)^2

DPatrick
2020-05-14 16:44:30

It's $(x-3)^2 + (y-4)^2.$

It's $(x-3)^2 + (y-4)^2.$

DPatrick
2020-05-14 16:44:37

And for the dart to be successful, we need the sum of these to be less than $30$.

And for the dart to be successful, we need the sum of these to be less than $30$.

DPatrick
2020-05-14 16:44:41

That is, we need $$\left(x^2+y^2\right) + \left((x-6)^2+y^2\right) + \left((x-3)^2+(y-4)^2\right) < 30.$$

That is, we need $$\left(x^2+y^2\right) + \left((x-6)^2+y^2\right) + \left((x-3)^2+(y-4)^2\right) < 30.$$

DPatrick
2020-05-14 16:44:50

Oof, what a mess. What should we do now?

Oof, what a mess. What should we do now?

Ninja-Girl
2020-05-14 16:45:15

It looks like a circle!

It looks like a circle!

aidni47
2020-05-14 16:45:15

that's in the form of the equation for a circle!

that's in the form of the equation for a circle!

Pleaseletmewin
2020-05-14 16:45:15

That’s a circle

That’s a circle

DPatrick
2020-05-14 16:45:30

It does look vaguely like the equation for a circle.

It does look vaguely like the equation for a circle.

DPatrick
2020-05-14 16:45:55

And we shouldn't be too surprised if the question expects us to have $\pi$ in our answer.

And we shouldn't be too surprised if the question expects us to have $\pi$ in our answer.

DPatrick
2020-05-14 16:46:13

So we want to make this inequality look more like the equation for a circle.

So we want to make this inequality look more like the equation for a circle.

EricShi1685
2020-05-14 16:46:26

simplify

simplify

nascar48
2020-05-14 16:46:26

expand

expand

Shmileyface
2020-05-14 16:46:26

expand

expand

menlo
2020-05-14 16:46:26

combine like terms

combine like terms

universeking
2020-05-14 16:46:26

expand!

expand!

DonutHole
2020-05-14 16:46:26

expand

expand

fasterthanlight
2020-05-14 16:46:26

expand

expand

Asterlan
2020-05-14 16:46:26

Expand

Expand

Kruxe
2020-05-14 16:46:26

combine like terms?

combine like terms?

bedwinprusik578
2020-05-14 16:46:26

expand

expand

sixoneeight
2020-05-14 16:46:26

Expand?

Expand?

look4pilot
2020-05-14 16:46:26

expand?

expand?

DPatrick
2020-05-14 16:46:48

Sure, let's clean it up by expanding all the squares and combining like terms.

Sure, let's clean it up by expanding all the squares and combining like terms.

DPatrick
2020-05-14 16:46:57

I'll save us a little time: expanding and simplifying the left-hand side gives \[

3x^2 -18x +3y^2 -8y + 61 < 30.\]

I'll save us a little time: expanding and simplifying the left-hand side gives \[

3x^2 -18x +3y^2 -8y + 61 < 30.\]

DPatrick
2020-05-14 16:47:19

Hmmm...still doesn't look quite like a circle to me.

Hmmm...still doesn't look quite like a circle to me.

DPatrick
2020-05-14 16:47:32

A circle's equation looks like $$(x-h)^2 + (y-k)^2 = r^2.$$

A circle's equation looks like $$(x-h)^2 + (y-k)^2 = r^2.$$

tumbleweed
2020-05-14 16:47:47

complete the square

complete the square

Puffer13
2020-05-14 16:47:47

complete the square

complete the square

EricShi1685
2020-05-14 16:47:47

complete the square

complete the square

Alculator11
2020-05-14 16:47:47

Now, let's complete the square

Now, let's complete the square

Math5K
2020-05-14 16:47:47

complete the square

complete the square

menlo
2020-05-14 16:47:47

complete the square!!!!

complete the square!!!!

MathNerd555
2020-05-14 16:47:47

Completing the square?

Completing the square?

Leonard_my_dude
2020-05-14 16:47:47

complete the square

complete the square

FearlessTaurus
2020-05-14 16:47:47

Complete squares

Complete squares

happyhari
2020-05-14 16:47:47

complete the square now

complete the square now

cat_maniac_8
2020-05-14 16:47:47

Complete the square?

Complete the square?

bobamilkcha
2020-05-14 16:47:47

complete the square

complete the square

DPatrick
2020-05-14 16:47:55

Aha, you said the magic words!

Aha, you said the magic words!

DPatrick
2020-05-14 16:47:59

We need to complete the square in both $x$ and $y.$

We need to complete the square in both $x$ and $y.$

ant08
2020-05-14 16:48:06

divdide by 3

divdide by 3

DarthMaul
2020-05-14 16:48:06

subtract 61

subtract 61

edjar
2020-05-14 16:48:06

subtract the 61

subtract the 61

Ilikeminecraft
2020-05-14 16:48:06

divide by 3

divide by 3

Lux1
2020-05-14 16:48:06

now we can subtract 61 from both sides

now we can subtract 61 from both sides

Ninja11
2020-05-14 16:48:09

divide by 3

divide by 3

rjiangbz
2020-05-14 16:48:09

divide by 3

divide by 3

jef23
2020-05-14 16:48:09

Subtract 61 from both sides

Subtract 61 from both sides

DPatrick
2020-05-14 16:48:21

Good idea: let's clean it up a bit more first by subtracting $61$ from both sides and dividing both sides by $3,$ to get \[

x^2 -6x + y^2 - \frac83y < -\frac{31}{3}.\]

Good idea: let's clean it up a bit more first by subtracting $61$ from both sides and dividing both sides by $3,$ to get \[

x^2 -6x + y^2 - \frac83y < -\frac{31}{3}.\]

DPatrick
2020-05-14 16:48:37

Now what do we need to do to complete the square for $x$?

Now what do we need to do to complete the square for $x$?

sosiaops
2020-05-14 16:48:56

+9

+9

evanshawn316
2020-05-14 16:48:56

add 9

add 9

agentmath
2020-05-14 16:48:56

add 9

add 9

PureSwag
2020-05-14 16:48:56

add $9$

add $9$

KevinW12345
2020-05-14 16:48:56

add 9

add 9

xMidnightFirex
2020-05-14 16:48:56

add 9

add 9

asbodke
2020-05-14 16:48:56

add 9 on both sides

add 9 on both sides

MathIsFun286
2020-05-14 16:48:56

add $9$

add $9$

pikachu5829
2020-05-14 16:48:56

+9

+9

Zhaom
2020-05-14 16:48:56

add 9

add 9

AMC_Kid
2020-05-14 16:48:56

add 9

add 9

DPatrick
2020-05-14 16:49:00

We need to add $9$ to complete the square for $x$, since $x^2 - 6x + 9 = (x-3)^2.$

We need to add $9$ to complete the square for $x$, since $x^2 - 6x + 9 = (x-3)^2.$

DPatrick
2020-05-14 16:49:07

And what about for $y$?

And what about for $y$?

Chesssaga
2020-05-14 16:49:22

and add 16/9

and add 16/9

Heavytoothpaste
2020-05-14 16:49:22

add 16/9 to both sides

add 16/9 to both sides

Jomo
2020-05-14 16:49:22

add 16/9

add 16/9

kdraganov
2020-05-14 16:49:22

add $\dfrac{16}{9}$

add $\dfrac{16}{9}$

MR_67
2020-05-14 16:49:22

+16/9

+16/9

RedFireTruck
2020-05-14 16:49:22

add 16/9

add 16/9

AmpharosX
2020-05-14 16:49:22

+16/9

+16/9

CHIPPER33
2020-05-14 16:49:22

add 16/9

add 16/9

Dalar25
2020-05-14 16:49:22

+16/9

+16/9

genius_007
2020-05-14 16:49:22

add 16/9

add 16/9

DPatrick
2020-05-14 16:49:27

We need to add $\dfrac{16}{9}$ to complete the square for $y,$ since $y^2 - \dfrac83y + \dfrac{16}{9} = \left(y - \dfrac43\right)^2.$

We need to add $\dfrac{16}{9}$ to complete the square for $y,$ since $y^2 - \dfrac83y + \dfrac{16}{9} = \left(y - \dfrac43\right)^2.$

DPatrick
2020-05-14 16:49:35

And of course, if we add these to the left side of the inequality, we have to add them to the right side too!

And of course, if we add these to the left side of the inequality, we have to add them to the right side too!

DPatrick
2020-05-14 16:49:43

So the right side becomes $-\dfrac{31}{3} + 9 + \dfrac{16}{9} = \dfrac{-93+81+16}{9} = \dfrac49.$

So the right side becomes $-\dfrac{31}{3} + 9 + \dfrac{16}{9} = \dfrac{-93+81+16}{9} = \dfrac49.$

DPatrick
2020-05-14 16:49:57

Therefore, completing the square in $x$ and $y$ gives\[

(x-3)^2 + \left(y-\frac43\right)^2 < \frac49.\]

Therefore, completing the square in $x$ and $y$ gives\[

(x-3)^2 + \left(y-\frac43\right)^2 < \frac49.\]

DPatrick
2020-05-14 16:50:09

Great! And what is this region?

Great! And what is this region?

IAmTheHazard
2020-05-14 16:50:36

It's a circle with radius $\frac{2}{3}$!

It's a circle with radius $\frac{2}{3}$!

winterrain01
2020-05-14 16:50:36

A circle centered at (3, 4/3) with radius 2/3

A circle centered at (3, 4/3) with radius 2/3

KevinW12345
2020-05-14 16:50:36

a circle with center (3, 4/3)

a circle with center (3, 4/3)

jacobwu
2020-05-14 16:50:36

THE RADIUS is 2/3

THE RADIUS is 2/3

yukrant1
2020-05-14 16:50:36

A circle with radius 2/3 and center (3,4/3)

A circle with radius 2/3 and center (3,4/3)

Puffer13
2020-05-14 16:50:36

a circle centered at (3,4/3) with radius 2/3

a circle centered at (3,4/3) with radius 2/3

ryanfu2008
2020-05-14 16:50:36

a circle with radius 2/3

a circle with radius 2/3

menlo
2020-05-14 16:50:36

center at (3,4/3)

center at (3,4/3)

phoenixtan
2020-05-14 16:50:36

center is 3,4/3, radius is 2/3

center is 3,4/3, radius is 2/3

Pokemon2
2020-05-14 16:50:36

that's a circle with center (3,4/3) and radius 2/3

that's a circle with center (3,4/3) and radius 2/3

DPatrick
2020-05-14 16:50:43

If this were an equation instead of an inequality, it would describe a circle with center $\left(3,\dfrac43\right)$ and radius $\sqrt{\dfrac49} = \dfrac23.$

If this were an equation instead of an inequality, it would describe a circle with center $\left(3,\dfrac43\right)$ and radius $\sqrt{\dfrac49} = \dfrac23.$

Eth007
2020-05-14 16:50:54

the inside of a circle

the inside of a circle

Facejo
2020-05-14 16:51:00

Inside a circle with radius $\frac{2}{3}$

Inside a circle with radius $\frac{2}{3}$

DPatrick
2020-05-14 16:51:03

So the inequality describes the interior of that circle, which is entirely inside the triangular dartboard.

So the inequality describes the interior of that circle, which is entirely inside the triangular dartboard.

DPatrick
2020-05-14 16:51:06

EulerRocks2.718
2020-05-14 16:51:32

Area of Success

Area of Success

dineshs
2020-05-14 16:51:32

disc with area 4pi/9

disc with area 4pi/9

mathfun42
2020-05-14 16:51:32

a circle with area 4pi/9

a circle with area 4pi/9

casp
2020-05-14 16:51:32

4pi/9

4pi/9

NonOxyCrustacean
2020-05-14 16:51:32

4/9pi

4/9pi

cai40
2020-05-14 16:51:32

so the successful area is 4/9pi

so the successful area is 4/9pi

Mandakini
2020-05-14 16:51:32

now just pi r^2

now just pi r^2

vsurya
2020-05-14 16:51:32

4/9 pi is area

4/9 pi is area

DPatrick
2020-05-14 16:51:37

It's area is $\pi r^2 = \pi\left(\dfrac23\right)^2 = \dfrac49\pi.$

It's area is $\pi r^2 = \pi\left(\dfrac23\right)^2 = \dfrac49\pi.$

DPatrick
2020-05-14 16:51:43

That's the area of our "successful region".

That's the area of our "successful region".

DPatrick
2020-05-14 16:51:49

And to finish, what is the probability?

And to finish, what is the probability?

mmjguitar
2020-05-14 16:52:11

pi/27

pi/27

ellnoo
2020-05-14 16:52:11

pi/27

pi/27

donguri
2020-05-14 16:52:11

So the answer is $\frac{\pi }{27}$

So the answer is $\frac{\pi }{27}$

slodha
2020-05-14 16:52:11

4/9 pi / 12

4/9 pi / 12

CT17
2020-05-14 16:52:11

$\frac{\frac{4}{9}\pi}{12}=\frac{\pi}{27}$

$\frac{\frac{4}{9}\pi}{12}=\frac{\pi}{27}$

yizhou76
2020-05-14 16:52:11

Our answer is 4pi/9/12=pi/27

Our answer is 4pi/9/12=pi/27

millburn2006
2020-05-14 16:52:11

pi/27

pi/27

Eng123
2020-05-14 16:52:11

$\pi/27$

$\pi/27$

onedance
2020-05-14 16:52:11

so it is pi/27

so it is pi/27

andrewhaitian
2020-05-14 16:52:11

pi/27

pi/27

slodha
2020-05-14 16:52:11

4 / 9 pi / 12 is the probability

4 / 9 pi / 12 is the probability

ZOP123
2020-05-14 16:52:11

pi/27

pi/27

DPatrick
2020-05-14 16:52:14

It is $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}} =\dfrac{\frac49\pi}{12}.$$

It is $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}} =\dfrac{\frac49\pi}{12}.$$

DPatrick
2020-05-14 16:52:36

And this simplifies to our final answer of $\boxed{\frac{\pi}{27}}.$

And this simplifies to our final answer of $\boxed{\frac{\pi}{27}}.$

DPatrick
2020-05-14 16:52:57

OK, back to rrusczyk! I'll be back later.

OK, back to rrusczyk! I'll be back later.

Meyzeek_Saveer
2020-05-14 16:53:22

RICHARD MY MANNNNN

RICHARD MY MANNNNN

Puffer13
2020-05-14 16:53:22

HI RICHARD RUSCZYK

HI RICHARD RUSCZYK

Speedstorm
2020-05-14 16:53:22

Hi Mr. Rusczyk

Hi Mr. Rusczyk

KRISHIV
2020-05-14 16:53:22

hi richard!!

hi richard!!

pi_is_3.14
2020-05-14 16:53:22

hi rruscyzk

hi rruscyzk

rrusczyk
2020-05-14 16:53:24

Hi!

Hi!

tacos4life123
2020-05-14 16:53:28

wait why are we changing back and fourth

wait why are we changing back and fourth

rrusczyk
2020-05-14 16:53:44

Because Dave and I are old, and get tired trying to keep up with all of you.

Because Dave and I are old, and get tired trying to keep up with all of you.

DPatrick
2020-05-14 16:53:55

Speak for yourself.

Speak for yourself.

rponda
2020-05-14 16:53:59

You're not that old!

You're not that old!

Pythongoras3
2020-05-14 16:54:22

how 'bout a break for us

how 'bout a break for us

rrusczyk
2020-05-14 16:54:31

Y'all don't look like you need a break to me!

Y'all don't look like you need a break to me!

rrusczyk
2020-05-14 16:54:43

You might after the next problem...

You might after the next problem...

rrusczyk
2020-05-14 16:54:45

**Sprint #30:**Hank builds an increasing sequence of positive integers as follows: The first term is $1$ and the second term is $2.$ Each subsequent term is the smallest positive integer that does NOT form a three-term arithmetic sequence with any previous terms of the sequence. The first five terms of Hank's sequence are $1, 2, 4, 5, 10.$ How many of the first $729$ positive integers are terms in Hank's sequence?
sri1priya
2020-05-14 16:55:03

i loved this problem!!

i loved this problem!!

rrusczyk
2020-05-14 16:55:06

Me, too!

Me, too!

rrusczyk
2020-05-14 16:55:07

Before we get too deep: is there anything suspicious in the problem statement?

Before we get too deep: is there anything suspicious in the problem statement?

edjar
2020-05-14 16:55:54

729

729

tenebrine
2020-05-14 16:55:54

729 = 3^6

729 = 3^6

walrus987
2020-05-14 16:55:54

729?

729?

AGENT99
2020-05-14 16:55:54

729

729

mathlete5451006
2020-05-14 16:55:54

729

729

raghavendrapv
2020-05-14 16:55:54

Why 729?

Why 729?

lamphead
2020-05-14 16:55:54

$729=3^6$

$729=3^6$

Flameslinger001
2020-05-14 16:55:54

why $729$

why $729$

Himank_Chhaya_3
2020-05-14 16:55:54

The 729

The 729

missionsqhc
2020-05-14 16:55:54

729 = 3 ^6

729 = 3 ^6

Noam2007
2020-05-14 16:55:54

Yes! $729=9^3=3^6.$

Yes! $729=9^3=3^6.$

rrusczyk
2020-05-14 16:55:58

The number $729$ is a little suspicious to me.

The number $729$ is a little suspicious to me.

rrusczyk
2020-05-14 16:55:59

After all, $729 = 3^6$ is a nice perfect power.

After all, $729 = 3^6$ is a nice perfect power.

rrusczyk
2020-05-14 16:56:21

So, we might have our eyes peeled for powers of 3.

So, we might have our eyes peeled for powers of 3.

rrusczyk
2020-05-14 16:56:42

Where should we start?

Where should we start?

evanshawn316
2020-05-14 16:56:54

patterns?

patterns?

Bluejay24
2020-05-14 16:56:54

the sequence?????

the sequence?????

MTHJJS
2020-05-14 16:56:54

lets find a pattern!

lets find a pattern!

Wontoflonto
2020-05-14 16:56:54

find a pattern?

find a pattern?

rrusczyk
2020-05-14 16:57:00

Let's look for some patterns.

Let's look for some patterns.

rrusczyk
2020-05-14 16:57:37

I see some of you asking what the question means. Let's figure that out by trying to build Hank's sequence. Maybe that will help us figure out what's going on this problem.

I see some of you asking what the question means. Let's figure that out by trying to build Hank's sequence. Maybe that will help us figure out what's going on this problem.

rrusczyk
2020-05-14 16:57:54

This "messing around" is a very powerful problem-solving strategy.

This "messing around" is a very powerful problem-solving strategy.

rrusczyk
2020-05-14 16:58:18

All right, let's start from the beginning:

All right, let's start from the beginning:

rrusczyk
2020-05-14 16:58:23

rrusczyk
2020-05-14 16:58:32

We include $1$.

We include $1$.

MLiang2018
2020-05-14 16:58:45

2

2

MLiang2018
2020-05-14 16:58:45

2

2

gs_2006
2020-05-14 16:58:45

next is 2

next is 2

MathNerd555
2020-05-14 16:58:45

and 2

and 2

xMidnightFirex
2020-05-14 16:58:45

and 2

and 2

akpi2
2020-05-14 16:58:52

second 2

second 2

Jing.Han
2020-05-14 16:58:52

2

2

cai40
2020-05-14 16:58:52

the second term is 2

the second term is 2

RedFireTruck
2020-05-14 16:58:52

2

2

rrusczyk
2020-05-14 16:59:00

Next is 2:

Next is 2:

rrusczyk
2020-05-14 16:59:04

rrusczyk
2020-05-14 16:59:11

Can we include the 3?

Can we include the 3?

qianqian07
2020-05-14 16:59:31

no

no

EricShi1685
2020-05-14 16:59:31

no

no

moony_eyed
2020-05-14 16:59:31

no

no

doudou_boston
2020-05-14 16:59:31

no

no

yesufsa
2020-05-14 16:59:31

No

No

mathissocool
2020-05-14 16:59:31

no

no

FALCONSRULE1
2020-05-14 16:59:31

no

no

vtlev
2020-05-14 16:59:31

no

no

FearlessTaurus
2020-05-14 16:59:31

no

no

Math_Genius_beast.com
2020-05-14 16:59:31

no

no

yayatheduck
2020-05-14 16:59:31

no

no

Ilikeminecraft
2020-05-14 16:59:31

no

no

Eng123
2020-05-14 16:59:31

NOOOOO.

NOOOOO.

EulerRocks2.718
2020-05-14 16:59:31

no

no

rneelam
2020-05-14 16:59:31

no

no

BurgerKingFootMan
2020-05-14 16:59:31

no

no

rrusczyk
2020-05-14 16:59:42

There were about 300 more no's in response...

There were about 300 more no's in response...

rrusczyk
2020-05-14 16:59:48

Why can't we include the 3?

Why can't we include the 3?

jacobwu
2020-05-14 17:00:00

NO that would make a sequence

NO that would make a sequence

CHIPPER33
2020-05-14 17:00:00

NO! 1,2,3 is an arithmetic sequence

NO! 1,2,3 is an arithmetic sequence

christopherp
2020-05-14 17:00:00

no, 1-2-3 is an arithmetic sequence

no, 1-2-3 is an arithmetic sequence

ant08
2020-05-14 17:00:00

no, 1+2=3

no, 1+2=3

Wontoflonto
2020-05-14 17:00:00

nah, 1-2-3 is a sequence

nah, 1-2-3 is a sequence

nmadhu
2020-05-14 17:00:00

No since 1, 2, 3 is an arithmetic sequence.

No since 1, 2, 3 is an arithmetic sequence.

rrusczyk
2020-05-14 17:00:41

Right -- 1,2,3 makes an arithmetic sequence: we add the same thing to get from the first to the second term as we do to get from the second to the third.

Right -- 1,2,3 makes an arithmetic sequence: we add the same thing to get from the first to the second term as we do to get from the second to the third.

rrusczyk
2020-05-14 17:00:43

So, no 3:

So, no 3:

rrusczyk
2020-05-14 17:00:46

MLiang2018
2020-05-14 17:01:03

so 4

so 4

Somersett
2020-05-14 17:01:03

Next is 4

Next is 4

Ninja11
2020-05-14 17:01:03

we can include 4

we can include 4

Alculator11
2020-05-14 17:01:03

But we can have 4

But we can have 4

ElNoraa
2020-05-14 17:01:03

4

4

akpi
2020-05-14 17:01:03

so 4

so 4

Mandakini
2020-05-14 17:01:03

then 4

then 4

Meyzeek_Saveer
2020-05-14 17:01:03

yes 4

yes 4

romani
2020-05-14 17:01:03

then 4

then 4

cheerupaops
2020-05-14 17:01:03

4

4

Moonshine-Dragonwing
2020-05-14 17:01:03

next term is 4

next term is 4

Poki
2020-05-14 17:01:03

4 would work

4 would work

Lux1
2020-05-14 17:01:03

then we have 4

then we have 4

dolphin7
2020-05-14 17:01:03

we can do 4

we can do 4

rrusczyk
2020-05-14 17:01:08

$4$ is a keeper:

$4$ is a keeper:

rrusczyk
2020-05-14 17:01:12

Supernova283
2020-05-14 17:01:25

Put in 5

Put in 5

eagle702
2020-05-14 17:01:25

Then 5

Then 5

s0arbeacon
2020-05-14 17:01:25

than 5

than 5

superagh
2020-05-14 17:01:25

5

5

karthic7073
2020-05-14 17:01:25

And then 5

And then 5

Kruxe
2020-05-14 17:01:25

then 5 is next

then 5 is next

rrusczyk
2020-05-14 17:01:45

Before we go circling $5,$ are there any numbers we know we can cross out now that we have included $4?$

Before we go circling $5,$ are there any numbers we know we can cross out now that we have included $4?$

Flameslinger001
2020-05-14 17:02:25

7

7

sarahAops2020
2020-05-14 17:02:25

7

7

fasterthanlight
2020-05-14 17:02:25

6

6

donguri
2020-05-14 17:02:25

6 and 7

6 and 7

DZL1
2020-05-14 17:02:25

6 and 7

6 and 7

punkinpiday
2020-05-14 17:02:25

6 and 7

6 and 7

rjiangbz
2020-05-14 17:02:25

6,7

6,7

pi_is_3.14
2020-05-14 17:02:25

6,7

6,7

enkitty
2020-05-14 17:02:25

We can't have 6 because of 4-5-6 and 2-4-6

We can't have 6 because of 4-5-6 and 2-4-6

ASMSRJ
2020-05-14 17:02:25

6,7

6,7

blueocean2019
2020-05-14 17:02:25

we can cross out 6

we can cross out 6

srim1027
2020-05-14 17:02:25

6, 7

6, 7

mathcounts_genius
2020-05-14 17:02:25

7

7

Flameslinger001
2020-05-14 17:02:25

6,7

6,7

Math_Penguin123
2020-05-14 17:02:25

7 and 6

7 and 6

Thegreatboy90
2020-05-14 17:02:25

6 and 7

6 and 7

AwesomeLife_Math
2020-05-14 17:02:25

7 and 6.

7 and 6.

rrusczyk
2020-05-14 17:02:30

We have the sequences $1$-$4$-$7$ and $2$-$4$-$6$.

We have the sequences $1$-$4$-$7$ and $2$-$4$-$6$.

rrusczyk
2020-05-14 17:02:34

So $6$ and $7$ get crossed out:

So $6$ and $7$ get crossed out:

rrusczyk
2020-05-14 17:02:37

rrusczyk
2020-05-14 17:03:21

This is pretty sweet. Once we find a new number, we can immediately cross others out.

This is pretty sweet. Once we find a new number, we can immediately cross others out.

tumbleweed
2020-05-14 17:03:32

we can still have 5

we can still have 5

Doudou_Chen
2020-05-14 17:03:32

and 5 gets circled?

and 5 gets circled?

Unicornzrock
2020-05-14 17:03:32

5 can be circleds

5 can be circleds

ZippityA
2020-05-14 17:03:32

5 is okay to circle.

5 is okay to circle.

Gumball
2020-05-14 17:03:35

we should circle 5

we should circle 5

superagh
2020-05-14 17:03:39

circle 5

circle 5

rrusczyk
2020-05-14 17:04:12

Now, we know that $5$ is a keeper because we have already eliminated all the ones that must be eliminated when we keep $1,2,$ and $4.$

Now, we know that $5$ is a keeper because we have already eliminated all the ones that must be eliminated when we keep $1,2,$ and $4.$

rrusczyk
2020-05-14 17:04:21

$5$ is the smallest unused number, and since we haven't crossed it out, it's not part of an arithmetic sequence. So we circle it.

$5$ is the smallest unused number, and since we haven't crossed it out, it's not part of an arithmetic sequence. So we circle it.

rrusczyk
2020-05-14 17:04:23

menlo
2020-05-14 17:04:58

8 & 9 cross them out

8 & 9 cross them out

Math4Life7
2020-05-14 17:04:58

now we can also cross out 8 and 9

now we can also cross out 8 and 9

kdraganov
2020-05-14 17:04:58

cross out 8 and 9 then

cross out 8 and 9 then

Juneybug
2020-05-14 17:04:58

so that eliminates 8 and 9 for us

so that eliminates 8 and 9 for us

Knightofaops
2020-05-14 17:04:58

cross out 8,9

cross out 8,9

JQWERTY6
2020-05-14 17:04:58

not nine.... 1-5-9 is a sequence

not nine.... 1-5-9 is a sequence

sixoneeight
2020-05-14 17:04:58

2-5-8, so 8 is crossed out, and 1-5-9, so 9 is crossed out.

2-5-8, so 8 is crossed out, and 1-5-9, so 9 is crossed out.

pandax2007
2020-05-14 17:04:58

8 doesnt work because 2-5-8

8 doesnt work because 2-5-8

Windigo
2020-05-14 17:04:58

no 8 or 9 because 1-5-9 and 2-5-8

no 8 or 9 because 1-5-9 and 2-5-8

MuffledPie
2020-05-14 17:04:58

then no 8 or 9

then no 8 or 9

mathfun42
2020-05-14 17:04:58

8 and 9 can be eliminated

8 and 9 can be eliminated

MathIsFun286
2020-05-14 17:04:58

cross out 8

cross out 8

evanshawn316
2020-05-14 17:04:58

we can eliminate 8 now that we have 5

we can eliminate 8 now that we have 5

slodha
2020-05-14 17:04:58

remove 9 and 8

remove 9 and 8

Quaoar
2020-05-14 17:04:58

That eliminates 8 and 9

That eliminates 8 and 9

btc433
2020-05-14 17:04:58

8 cannot work because 2,5,8 is an arithmetic sequence now that 5 is circled.

8 cannot work because 2,5,8 is an arithmetic sequence now that 5 is circled.

KingRavi
2020-05-14 17:04:58

cross out 8 and 9

cross out 8 and 9

mfro24
2020-05-14 17:04:58

now cross out 8 and 9

now cross out 8 and 9

jai123
2020-05-14 17:04:58

then take out 8 and 9 1-5-9 and 2-5-8

then take out 8 and 9 1-5-9 and 2-5-8

Makemakika1
2020-05-14 17:04:58

No 8 (2, 5, 8), no 9 (1, 5, 9)

No 8 (2, 5, 8), no 9 (1, 5, 9)

agentmath
2020-05-14 17:04:58

8 doesn't work because 2, 5, 8, 9 doesn't work because of 1, 5, 9

8 doesn't work because 2, 5, 8, 9 doesn't work because of 1, 5, 9

LovingPilot
2020-05-14 17:04:58

not 8 and 9, 2-5-8, 1-5,9

not 8 and 9, 2-5-8, 1-5,9

rrusczyk
2020-05-14 17:05:20

The sequences with 5 in the middle are:

1-5-9

2-5-8

4-5-6

The sequences with 5 in the middle are:

1-5-9

2-5-8

4-5-6

rrusczyk
2020-05-14 17:05:23

So we have to cross out $8$ and $9$. ($6$ is already crossed-out.)

So we have to cross out $8$ and $9$. ($6$ is already crossed-out.)

rrusczyk
2020-05-14 17:05:24

rrusczyk
2020-05-14 17:06:04

This is fun -- it's a little like the Sieve of Eratosthenes for finding primes!

This is fun -- it's a little like the Sieve of Eratosthenes for finding primes!

doglover07
2020-05-14 17:06:26

circle 10!

circle 10!

HarleyMathCounts
2020-05-14 17:06:26

circle 10

circle 10

kkomma10
2020-05-14 17:06:26

you can do 10

you can do 10

Heavytoothpaste
2020-05-14 17:06:26

10 works

10 works

mathking999
2020-05-14 17:06:26

We can circle 10.

We can circle 10.

smartatmath
2020-05-14 17:06:26

10 circle

10 circle

palindrome868
2020-05-14 17:06:26

We can add 10

We can add 10

Chesssaga
2020-05-14 17:06:26

10 is a keeper

10 is a keeper

Anish.A
2020-05-14 17:06:26

10 does work

10 does work

artemispi
2020-05-14 17:06:26

circle 10

circle 10

ttmmrryy
2020-05-14 17:06:26

next is 10!!

next is 10!!

gs_2006
2020-05-14 17:06:26

then circle the 10

then circle the 10

anishcool11
2020-05-14 17:06:26

circle 10

circle 10

rrusczyk
2020-05-14 17:06:31

We circle $10$, the smallest unmarked numbered:

We circle $10$, the smallest unmarked numbered:

rrusczyk
2020-05-14 17:06:32

Lux1
2020-05-14 17:07:16

and cross out 19, 18, 16, 15

and cross out 19, 18, 16, 15

zhuqingzhang
2020-05-14 17:07:16

once we put in 10, we can eliminate 15,16,18,and 19

once we put in 10, we can eliminate 15,16,18,and 19

nduracell
2020-05-14 17:07:16

Once we circle 10, we can cross off 15, 16, 18 and 19

Once we circle 10, we can cross off 15, 16, 18 and 19

jbear911
2020-05-14 17:07:16

Cross out 15 because 5-10-15

Cross out 15 because 5-10-15

mathfun42
2020-05-14 17:07:16

You can cross out 15, 16, 18, and 19

You can cross out 15, 16, 18, and 19

edjar
2020-05-14 17:07:16

Cross out 15,16,18,19

Cross out 15,16,18,19

bluemathcounts
2020-05-14 17:07:16

cross out 15,16,18, and 19

cross out 15,16,18, and 19

flissyquokka17
2020-05-14 17:07:16

then eliminate 19, 18, 16, 15: (1, 10, 19), (2, 10, 18), (4, 10, 16), (5, 10, 15)

then eliminate 19, 18, 16, 15: (1, 10, 19), (2, 10, 18), (4, 10, 16), (5, 10, 15)

bluemathcounts
2020-05-14 17:07:16

cross out 15,16,18, and 19.

cross out 15,16,18, and 19.

cj13609517288
2020-05-14 17:07:16

15,16,18,19 are out.

15,16,18,19 are out.

melonlord
2020-05-14 17:07:16

cross out 15,16,18, and 19

cross out 15,16,18, and 19

coldcrazylogic
2020-05-14 17:07:16

15, 16, 18, and 19 are gone because of this

15, 16, 18, and 19 are gone because of this

nduracell
2020-05-14 17:07:16

We can cross off 15, 16 18 and 19

We can cross off 15, 16 18 and 19

rrusczyk
2020-05-14 17:07:40

All the numbers smaller than $10$ can form a sequence with $10$ like so:

1-10-19

2-10-18

4-10-16

5-10-15

All the numbers smaller than $10$ can form a sequence with $10$ like so:

1-10-19

2-10-18

4-10-16

5-10-15

rrusczyk
2020-05-14 17:07:42

So $15$, $16$, $18$, and $19$ all get crossed out.

So $15$, $16$, $18$, and $19$ all get crossed out.

rrusczyk
2020-05-14 17:07:44

nascar48
2020-05-14 17:08:01

11 works

11 works

broham
2020-05-14 17:08:01

11 works

11 works

sarahAops2020
2020-05-14 17:08:01

11

11

truffle
2020-05-14 17:08:01

11 works too

11 works too

LlamaWarrior
2020-05-14 17:08:01

Circle 11 as well

Circle 11 as well

Unicorn78
2020-05-14 17:08:01

I think 11 works too

I think 11 works too

halphaboys
2020-05-14 17:08:01

circle 11

circle 11

Poki
2020-05-14 17:08:05

then 11 would work also

then 11 would work also

rrusczyk
2020-05-14 17:08:21

We circle 11, and we cross out:

1-11-21

2-11-20

4-11-18

5-11-17

10-11-12

We circle 11, and we cross out:

1-11-21

2-11-20

4-11-18

5-11-17

10-11-12

rrusczyk
2020-05-14 17:08:23

Facejo
2020-05-14 17:08:49

circle $13$

circle $13$

kkomma10
2020-05-14 17:08:49

you can do 13

you can do 13

JCanada
2020-05-14 17:08:49

Then 13

Then 13

donguri
2020-05-14 17:08:49

circle 13

circle 13

MTHJJS
2020-05-14 17:08:49

and 13 works

and 13 works

sdattilo2002
2020-05-14 17:08:49

13 works

13 works

Thegreatboy90
2020-05-14 17:08:49

13 is next

13 is next

MathNerd555
2020-05-14 17:08:49

13 works

13 works

MathMaster029
2020-05-14 17:08:49

13 works, circle it

13 works, circle it

rrusczyk
2020-05-14 17:08:53

We can continue like this.

We can continue like this.

rrusczyk
2020-05-14 17:08:54

$13$ gets circled, and $25$, $24$, $22$, $21$ will get crossed out.

$13$ gets circled, and $25$, $24$, $22$, $21$ will get crossed out.

rrusczyk
2020-05-14 17:08:55

Layla2018
2020-05-14 17:09:18

circle 14

circle 14

kittysnowball43
2020-05-14 17:09:18

14 works too

14 works too

melonlord
2020-05-14 17:09:18

crcle 14

crcle 14

ElNoraa
2020-05-14 17:09:18

14

14

donguri
2020-05-14 17:09:18

circle 14

circle 14

bot101
2020-05-14 17:09:18

14

14

yesufsa
2020-05-14 17:09:18

14 can be circled

14 can be circled

walrus987
2020-05-14 17:09:18

14 can stay

14 can stay

winterrain01
2020-05-14 17:09:18

Circle 14

Circle 14

happyhari
2020-05-14 17:09:18

then 14

then 14

danprathab
2020-05-14 17:09:18

Circle 14

Circle 14

The_math_kid
2020-05-14 17:09:18

14 works

14 works

rrusczyk
2020-05-14 17:09:20

$14$ gets circled, and $27$, $26$, and $23$ get crossed out.

$14$ gets circled, and $27$, $26$, and $23$ get crossed out.

rrusczyk
2020-05-14 17:09:21

rrusczyk
2020-05-14 17:09:35

So the next number in the sequence is $28$.

So the next number in the sequence is $28$.

jef23
2020-05-14 17:09:59

Except we can’t just go on like this till 729...

Except we can’t just go on like this till 729...

doglover07
2020-05-14 17:09:59

this is gonna take forever

this is gonna take forever

rrusczyk
2020-05-14 17:10:11

Yeah, this is gonna take a while if we keep going...

Yeah, this is gonna take a while if we keep going...

rrusczyk
2020-05-14 17:10:30

Now, back at the beginning, what did we decide we were going to look out for?

Now, back at the beginning, what did we decide we were going to look out for?

FearlessTaurus
2020-05-14 17:10:47

powers of 3

powers of 3

aidni47
2020-05-14 17:10:47

powers of 3

powers of 3

Green4Applez
2020-05-14 17:10:47

powers of 3

powers of 3

BurgerKingFootMan
2020-05-14 17:10:47

powers of 3!

powers of 3!

menlo
2020-05-14 17:10:47

multiles/powers of 3

multiles/powers of 3

scoutskylar
2020-05-14 17:10:47

powers of 3

powers of 3

onedance
2020-05-14 17:10:50

powers of 3

powers of 3

harmonyguan
2020-05-14 17:10:50

powers of 3

powers of 3

rrusczyk
2020-05-14 17:10:57

All right, let's look at our grid.

All right, let's look at our grid.

rrusczyk
2020-05-14 17:11:13

AGENT99
2020-05-14 17:11:25

All the numbers after powers of 3 work

All the numbers after powers of 3 work

vtlev
2020-05-14 17:11:29

10=3^2+1 and 28=3^3+1

10=3^2+1 and 28=3^3+1

cj13609517288
2020-05-14 17:11:32

10=3^2+1 and 28=3^3+1

10=3^2+1 and 28=3^3+1

rrusczyk
2020-05-14 17:11:40

Oh, that's interesting!

Oh, that's interesting!

punkinpiday
2020-05-14 17:12:02

2 of the first 3, 4 of the first 9, 8 of the first 27.. 64 of the first 729!

2 of the first 3, 4 of the first 9, 8 of the first 27.. 64 of the first 729!

john0512
2020-05-14 17:12:02

2 circled to 3, 4 circled to 9, 8 circled to 27

2 circled to 3, 4 circled to 9, 8 circled to 27

scoutskylar
2020-05-14 17:12:14

Up to 3^1, there are 2^1 numbers in the sequence. Up to 3^2, there are 2^2. Up to 3^3, there are 2^3. This pattern continues: up to 3^n, there are 2^n numbers in the sequence.

Up to 3^1, there are 2^1 numbers in the sequence. Up to 3^2, there are 2^2. Up to 3^3, there are 2^3. This pattern continues: up to 3^n, there are 2^n numbers in the sequence.

rrusczyk
2020-05-14 17:12:17

Ah, that's interesting, too!

Ah, that's interesting, too!

samsonrao
2020-05-14 17:12:54

2 numbers before 3, 4 numbers before 9, 8 numbers before 27. Maybe 2^n numbers before 3^n

2 numbers before 3, 4 numbers before 9, 8 numbers before 27. Maybe 2^n numbers before 3^n

rrusczyk
2020-05-14 17:13:08

So, we see something interesting in our powers of $3.$ And some of us are seeing powers of $2,$ too. Let's stick with the powers of $3$ first.

So, we see something interesting in our powers of $3.$ And some of us are seeing powers of $2,$ too. Let's stick with the powers of $3$ first.

rrusczyk
2020-05-14 17:13:34

How might we look at each integer if we are thinking about powers of $3?$

How might we look at each integer if we are thinking about powers of $3?$

Pleaseletmewin
2020-05-14 17:13:55

put them in base 3

put them in base 3

Speedstorm
2020-05-14 17:13:55

Base 3!

Base 3!

CHIPPER33
2020-05-14 17:13:55

base 3

base 3

pi_is_3.14
2020-05-14 17:13:55

base 3

base 3

dolphin7
2020-05-14 17:13:55

base 3

base 3

amuthupss
2020-05-14 17:13:55

in base 3?

in base 3?

Speedstorm
2020-05-14 17:13:55

base 3

base 3

rrusczyk
2020-05-14 17:14:10

Lot's of stuff about powers of $3,$ let's think in base $3.$

Lot's of stuff about powers of $3,$ let's think in base $3.$

rrusczyk
2020-05-14 17:14:20

Hank's sequence so far then would be:

Hank's sequence so far then would be:

rrusczyk
2020-05-14 17:14:28

$$1_3, 2_3, 11_3, 12_3, 101_3, 102_3, 111_3, 112_3, 1001_3.$$

$$1_3, 2_3, 11_3, 12_3, 101_3, 102_3, 111_3, 112_3, 1001_3.$$

mathking999
2020-05-14 17:14:47

All the ending units digits are one or two

All the ending units digits are one or two

Robot7620_2
2020-05-14 17:15:07

everything ends in 1 or 2

everything ends in 1 or 2

donguri
2020-05-14 17:15:07

Units digit are either 1 or 2

Units digit are either 1 or 2

Jomo
2020-05-14 17:15:07

All the ones that have first digits of 1s and last digits of 1 or 2

All the ones that have first digits of 1s and last digits of 1 or 2

rrusczyk
2020-05-14 17:15:11

Indeed, the final digit is always $1$ or $2.$

Indeed, the final digit is always $1$ or $2.$

rrusczyk
2020-05-14 17:15:15

And the other digits?

And the other digits?

edjar
2020-05-14 17:15:37

and all the rest are 0 or 1

and all the rest are 0 or 1

mfro24
2020-05-14 17:15:37

First two digits are all 0 or 1

First two digits are all 0 or 1

mfro24
2020-05-14 17:15:37

always 1 or 0

always 1 or 0

Superior
2020-05-14 17:15:37

1 and 0s

1 and 0s

Gentoo
2020-05-14 17:15:37

1 or 0

1 or 0

edjar
2020-05-14 17:15:37

middles are 0 or 1

middles are 0 or 1

Alculator11
2020-05-14 17:15:39

The other digits are 1 or 0

The other digits are 1 or 0

ant08
2020-05-14 17:15:42

1 or 0

1 or 0

egerns
2020-05-14 17:15:42

0 or 1

0 or 1

olive0827
2020-05-14 17:15:42

just 1 or 0

just 1 or 0

nkt2002
2020-05-14 17:15:42

1 or 0

1 or 0

rrusczyk
2020-05-14 17:15:56

Just $0$ and $1$ for the other digits...

Just $0$ and $1$ for the other digits...

scoutskylar
2020-05-14 17:16:23

There aren't any digit 2s if you subtract 1!

There aren't any digit 2s if you subtract 1!

rrusczyk
2020-05-14 17:16:45

Oh, very interesting! Let's subtract $1$ from every number. Then we get:

Oh, very interesting! Let's subtract $1$ from every number. Then we get:

rrusczyk
2020-05-14 17:16:50

$$0_3, 1_3, 10_3, 11_3, 100_3, 101_3, 110_3, 111_3, 1000_3.$$

$$0_3, 1_3, 10_3, 11_3, 100_3, 101_3, 110_3, 111_3, 1000_3.$$

scinderella220
2020-05-14 17:17:00

Wow

Wow

Quaoar
2020-05-14 17:17:00

Woah.

Woah.

danprathab
2020-05-14 17:17:27

WOW

WOW

agentmath
2020-05-14 17:17:27

wow

wow

winterrain01
2020-05-14 17:17:27

Those look like binary

Those look like binary

LlamaWarrior
2020-05-14 17:17:27

Amazing

Amazing

Imayormaynotknowcalculus
2020-05-14 17:17:27

Pretty cool

Pretty cool

zhuqingzhang
2020-05-14 17:17:27

Wow

Wow

PShucks
2020-05-14 17:17:27

WOah1

WOah1

RedFireTruck
2020-05-14 17:17:27

thats binary

thats binary

r0518
2020-05-14 17:17:27

now we only have 0 and 1

now we only have 0 and 1

sri1priya
2020-05-14 17:17:27

it becomes binary!

it becomes binary!

EulerRocks2.718
2020-05-14 17:17:27

Binary

Binary

Ajtyb12dfj
2020-05-14 17:17:27

wow!

wow!

MathBluebird
2020-05-14 17:17:27

That's so cool!

That's so cool!

PShucks
2020-05-14 17:17:27

WOOAH!

WOOAH!

rrusczyk
2020-05-14 17:17:56

Yeah, that's just plain pretty. Let's just look at it for a little while.

Yeah, that's just plain pretty. Let's just look at it for a little while.

scoutskylar
2020-05-14 17:18:05

That isn't binary.

That isn't binary.

rrusczyk
2020-05-14 17:18:22

Correct -- we aren't looking at this in base 2 right now.

Correct -- we aren't looking at this in base 2 right now.

aidni47
2020-05-14 17:18:27

its in base 3

its in base 3

KRISHIV
2020-05-14 17:18:33

yea its in base 3 still

yea its in base 3 still

jef23
2020-05-14 17:18:33

Yeah bc it’s in base 3

Yeah bc it’s in base 3

BurgerKingFootMan
2020-05-14 17:18:33

that's base 3

that's base 3

rrusczyk
2020-05-14 17:18:36

It is in base 3.

It is in base 3.

rrusczyk
2020-05-14 17:18:39

But...

But...

dineshs
2020-05-14 17:18:41

It looks like binary

It looks like binary

rrusczyk
2020-05-14 17:18:53

It sure looks like binary.

It sure looks like binary.

fasterthanlight
2020-05-14 17:19:17

think of it as binary

think of it as binary

NHatAOPS
2020-05-14 17:19:17

if it was binary, it would be 1 2 3 4 5...

if it was binary, it would be 1 2 3 4 5...

floatmeeting
2020-05-14 17:19:17

It's every number in binary too

It's every number in binary too

rrusczyk
2020-05-14 17:20:06

If we just think of these as being binary numbers.... WE HAVE ALL THE POSITIVE INTEGERS. Which makes counting the number of terms in Hank's sequence up through a given number really easy.

If we just think of these as being binary numbers.... WE HAVE ALL THE POSITIVE INTEGERS. Which makes counting the number of terms in Hank's sequence up through a given number really easy.

nathanqiu
2020-05-14 17:20:13

AND THAT EXPLAINS THE 2^N! BECAUSE THERE ARE 2^N NUMBERS OF N DIGITS IN BINARY!!!

AND THAT EXPLAINS THE 2^N! BECAUSE THERE ARE 2^N NUMBERS OF N DIGITS IN BINARY!!!

Neberg
2020-05-14 17:20:21

There are two options for each digit, 0 or 1, so there are $2^n$ numbers less than $3^n$ (because there are 2 options for each of n digits).

There are two options for each digit, 0 or 1, so there are $2^n$ numbers less than $3^n$ (because there are 2 options for each of n digits).

rrusczyk
2020-05-14 17:20:22

Exactly.

Exactly.

rrusczyk
2020-05-14 17:20:45

So, we have a guess now for our answer.

So, we have a guess now for our answer.

MuffledPie
2020-05-14 17:21:09

729=3^6,so 2^6=64

729=3^6,so 2^6=64

Ninja11
2020-05-14 17:21:09

2^6 is 64

2^6 is 64

Chesssaga
2020-05-14 17:21:09

So the answer is 64

So the answer is 64

DottedCaculator
2020-05-14 17:21:09

64

64

cooljoseph
2020-05-14 17:21:09

64

64

try11out
2020-05-14 17:21:09

64

64

Windigo
2020-05-14 17:21:09

64

64

NHatAOPS
2020-05-14 17:21:09

64

64

MathWizard09
2020-05-14 17:21:09

64

64

manokid
2020-05-14 17:21:09

64

64

MathJams
2020-05-14 17:21:09

64?

64?

rrusczyk
2020-05-14 17:21:26

We think that there are $64$ terms in Hank's sequence.

We think that there are $64$ terms in Hank's sequence.

rrusczyk
2020-05-14 17:21:33

Let's recap why we believe this.

Let's recap why we believe this.

rrusczyk
2020-05-14 17:21:38

Since $729 = 1000000_3,$ the numbers from $0$ to $728$ are all the numbers we can write in base $3$ with $6$ or fewer digits.

Since $729 = 1000000_3,$ the numbers from $0$ to $728$ are all the numbers we can write in base $3$ with $6$ or fewer digits.

rrusczyk
2020-05-14 17:22:19

But in our list, we want only the numbers of $0$ or $1$ as digits. (Including our leading $0$s).

But in our list, we want only the numbers of $0$ or $1$ as digits. (Including our leading $0$s).

rrusczyk
2020-05-14 17:23:00

So, we have $6$ slots to fill (again, including possibly with those leading $0$s), and each slot has $2$ choices: $0$ or $1.$

So, we have $6$ slots to fill (again, including possibly with those leading $0$s), and each slot has $2$ choices: $0$ or $1.$

skyleristhecoolest
2020-05-14 17:23:05

so 2^6

so 2^6

AbhiWwis
2020-05-14 17:23:05

2^6

2^6

rrusczyk
2020-05-14 17:23:31

So, there are $2^6$ of these base-$3$ numbers less than $729$ with only $0$ and $1$ as digits.

So, there are $2^6$ of these base-$3$ numbers less than $729$ with only $0$ and $1$ as digits.

rrusczyk
2020-05-14 17:23:47

(That's another way to look at the observation we made by reading everything as binary.)

(That's another way to look at the observation we made by reading everything as binary.)

sri1priya
2020-05-14 17:23:59

that makes this problem the coolest!

that makes this problem the coolest!

rrusczyk
2020-05-14 17:24:02

Indeed.

Indeed.

rrusczyk
2020-05-14 17:24:04

But.

But.

Scipow
2020-05-14 17:24:18

but why does this happen?

but why does this happen?

cj13609517288
2020-05-14 17:24:18

Unfortunately, we need to prove this conjecture.

Unfortunately, we need to prove this conjecture.

Chesssaga
2020-05-14 17:24:27

we still need to prove it rigorously

we still need to prove it rigorously

amuthupss
2020-05-14 17:24:29

we need to prove it works

we need to prove it works

MR_67
2020-05-14 17:24:42

how do we prove this

how do we prove this

RedFireTruck
2020-05-14 17:24:42

we need to use a proof

we need to use a proof

rrusczyk
2020-05-14 17:24:46

Yeah, I'm not satisfied either.

Yeah, I'm not satisfied either.

rrusczyk
2020-05-14 17:24:55

I think we have the answer.

I think we have the answer.

rrusczyk
2020-05-14 17:25:03

But as a mathematician, I want to try to

But as a mathematician, I want to try to

**prove**that the conjecture is true.
rrusczyk
2020-05-14 17:25:23

(I might not do this during the test. But afterwards, wouldn't be able to sleep until I figured it out.)

(I might not do this during the test. But afterwards, wouldn't be able to sleep until I figured it out.)

Zoobat
2020-05-14 17:25:30

How do you do this in less than 2 minutes?

How do you do this in less than 2 minutes?

rrusczyk
2020-05-14 17:25:45

See the pattern, hope it continues, write down $64$, prove it later.

See the pattern, hope it continues, write down $64$, prove it later.

rneelam
2020-05-14 17:25:53

are we gonna prove it now

are we gonna prove it now

rrusczyk
2020-05-14 17:25:57

We're gonna try.

We're gonna try.

rrusczyk
2020-05-14 17:26:01

There are two steps: we have to prove that the sequence we have satisfies the "no three terms form an arithmetic sequence" property, and we have to prove that we can't add any new terms to the sequence to make it bigger.

There are two steps: we have to prove that the sequence we have satisfies the "no three terms form an arithmetic sequence" property, and we have to prove that we can't add any new terms to the sequence to make it bigger.

rrusczyk
2020-05-14 17:26:15

Let's do these one at a time.

Let's do these one at a time.

rrusczyk
2020-05-14 17:26:30

First, we show that if we only have numbers with $0$s and $1$s as base-$3$ digits, then no three of them can form an arithmetic sequence.

First, we show that if we only have numbers with $0$s and $1$s as base-$3$ digits, then no three of them can form an arithmetic sequence.

rrusczyk
2020-05-14 17:26:37

Let's suppose we had three terms $x < y < z$. What equation would have to be true for this to be an arithmetic sequence?

Let's suppose we had three terms $x < y < z$. What equation would have to be true for this to be an arithmetic sequence?

Puffer13
2020-05-14 17:27:08

2y=x+z

2y=x+z

rjiangbz
2020-05-14 17:27:08

x+z=2y

x+z=2y

superagh
2020-05-14 17:27:08

2y=x+z

2y=x+z

doglover07
2020-05-14 17:27:08

y-x=z-y

y-x=z-y

XTJin
2020-05-14 17:27:08

x+z=2y

x+z=2y

Leonard_my_dude
2020-05-14 17:27:08

x+z=2y

x+z=2y

BakedPotato66
2020-05-14 17:27:08

x+z=2y

x+z=2y

cindyzou
2020-05-14 17:27:08

y-x = z-y

y-x = z-y

MY-2
2020-05-14 17:27:08

y-x = z-y

y-x = z-y

amuthupss
2020-05-14 17:27:08

y-x = z-y

y-x = z-y

edjar
2020-05-14 17:27:08

y-x=z-y

y-x=z-y

ant08
2020-05-14 17:27:08

x+z=2y

x+z=2y

MathNerd555
2020-05-14 17:27:08

y-x=z-y

y-x=z-y

rrusczyk
2020-05-14 17:27:31

We would need $y$ to be the average of $x$ and $z$. (Or we can start from $y-x = z-y,$ as many of you nicely did.)

We would need $y$ to be the average of $x$ and $z$. (Or we can start from $y-x = z-y,$ as many of you nicely did.)

rrusczyk
2020-05-14 17:27:36

The simplest way to write this is $2y = x+z$.

The simplest way to write this is $2y = x+z$.

rrusczyk
2020-05-14 17:28:56

Now, if $x, y,$ and $z$ are in our sequence written in base $3$ (after we subtracted $1$), then what do we know about that right-hand side $x+z?$

Now, if $x, y,$ and $z$ are in our sequence written in base $3$ (after we subtracted $1$), then what do we know about that right-hand side $x+z?$

donguri
2020-05-14 17:29:08

There are no carries?

There are no carries?

edjar
2020-05-14 17:29:08

No carrying

No carrying

acornfirst
2020-05-14 17:29:20

it would have no carries

it would have no carries

rrusczyk
2020-05-14 17:29:51

For each digit in the sum, we add a $0$ or $1$ to another $0$ or $1.$

For each digit in the sum, we add a $0$ or $1$ to another $0$ or $1.$

awesomebooks
2020-05-14 17:29:57

no carries

no carries

CHIPPER33
2020-05-14 17:29:57

no carries

no carries

john0512
2020-05-14 17:29:57

No carrying

No carrying

happyhari
2020-05-14 17:29:57

no carries

no carries

mohanty
2020-05-14 17:29:57

no carry over

no carry over

tigerjade003
2020-05-14 17:29:57

NO CARRIES

NO CARRIES

rrusczyk
2020-05-14 17:30:01

So, no carries.

So, no carries.

bishope6
2020-05-14 17:30:09

It could end in a 0, 1, or 2.

It could end in a 0, 1, or 2.

Green4Applez
2020-05-14 17:30:17

it will have digits of 0, 1, and 2

it will have digits of 0, 1, and 2

Maths4J
2020-05-14 17:30:17

It has 0s, 1s, and 2s in it

It has 0s, 1s, and 2s in it

rrusczyk
2020-05-14 17:30:30

Each digit will be $0,1,$ or $2.$ Hmmm....

Each digit will be $0,1,$ or $2.$ Hmmm....

rrusczyk
2020-05-14 17:30:38

Maybe that's not so interesting. Let's look at the other side.

Maybe that's not so interesting. Let's look at the other side.

rrusczyk
2020-05-14 17:30:44

What will the left side look like?

What will the left side look like?

Leonard_my_dude
2020-05-14 17:31:11

also no carries

also no carries

Chesssaga
2020-05-14 17:31:11

0 or 2

0 or 2

mfro24
2020-05-14 17:31:11

all 0s or 2s

all 0s or 2s

eez
2020-05-14 17:31:11

a series of 0s and 2s

a series of 0s and 2s

CHIPPER33
2020-05-14 17:31:11

end in 0 or 2

end in 0 or 2

AGENT99
2020-05-14 17:31:11

0 or 2

0 or 2

scoutskylar
2020-05-14 17:31:11

there can't be any 1s.

there can't be any 1s.

cj13609517288
2020-05-14 17:31:11

Each digit can only be 0 or 2.

Each digit can only be 0 or 2.

Eng123
2020-05-14 17:31:11

0 or 2

0 or 2

MathIsFun286
2020-05-14 17:31:11

no carries

no carries

FireDragon1719
2020-05-14 17:31:13

it will be either a $0,2$

it will be either a $0,2$

rrusczyk
2020-05-14 17:31:18

Also no carries.

Also no carries.

rrusczyk
2020-05-14 17:31:32

$y$ only has $0$ and $1$ as digits, so we double $y$ by doubling each digit.

$y$ only has $0$ and $1$ as digits, so we double $y$ by doubling each digit.

rrusczyk
2020-05-14 17:31:47

But there are no $1$s in $2y.$

But there are no $1$s in $2y.$

rrusczyk
2020-05-14 17:32:02

Looking back over at the right hand side....

Looking back over at the right hand side....

rrusczyk
2020-05-14 17:32:29

Is it possible that the right-hand side will not have any $1$s?

Is it possible that the right-hand side will not have any $1$s?

edjar
2020-05-14 17:32:59

And in the original there has to be at least 1 one

And in the original there has to be at least 1 one

astumbur
2020-05-14 17:32:59

no

no

PShucks
2020-05-14 17:32:59

no

no

maxben
2020-05-14 17:32:59

no

no

waddlethegreat
2020-05-14 17:32:59

no

no

PShucks
2020-05-14 17:32:59

no!

no!

cat_maniac_8
2020-05-14 17:32:59

No

No

m_goli
2020-05-14 17:32:59

no

no

DottedCaculator
2020-05-14 17:32:59

no

no

PShucks
2020-05-14 17:32:59

no!

no!

MathWizard09
2020-05-14 17:32:59

no

no

rrusczyk
2020-05-14 17:33:06

Why not?

Why not?

asbodke
2020-05-14 17:33:16

Only if they are the same number

Only if they are the same number

nathanqiu
2020-05-14 17:33:16

only if there is no 1+0 or 0+1

only if there is no 1+0 or 0+1

Gentoo
2020-05-14 17:33:16

Each 1 will have to match up with another 1

Each 1 will have to match up with another 1

mohanty
2020-05-14 17:33:33

there has to be at least one 1

there has to be at least one 1

tigerjade003
2020-05-14 17:33:33

there has to be at least one 1

there has to be at least one 1

rrusczyk
2020-05-14 17:33:42

Ah! $x$ and $z$ are different!

Ah! $x$ and $z$ are different!

rrusczyk
2020-05-14 17:33:54

Some digit of our sum $x+z$ will come from adding a $0$ from $x$ and a $1$ for $z,$ or the reverse.

Some digit of our sum $x+z$ will come from adding a $0$ from $x$ and a $1$ for $z,$ or the reverse.

Pleaseletmewin
2020-05-14 17:34:07

There has to be at least 1

There has to be at least 1

rrusczyk
2020-05-14 17:34:09

So, we have to get a $1$ somewhere!

So, we have to get a $1$ somewhere!

yukrant1
2020-05-14 17:34:31

then 2y can never equal x+z

then 2y can never equal x+z

rrusczyk
2020-05-14 17:34:37

So, $2y$ has no $1$s. But $x+z$ has at least one $1$.

So, $2y$ has no $1$s. But $x+z$ has at least one $1$.

Puffer13
2020-05-14 17:34:41

so they cannot be equal

so they cannot be equal

rrusczyk
2020-05-14 17:35:03

So we cannot have $2y = x + z$. And thus there's no arithmetic sequence in our sequence that has only those numbers with $0$s and $1$s.

So we cannot have $2y = x + z$. And thus there's no arithmetic sequence in our sequence that has only those numbers with $0$s and $1$s.

cj13609517288
2020-05-14 17:35:09

And we are done!!

Oh wait, that's only half of the proof.

And we are done!!

Oh wait, that's only half of the proof.

rrusczyk
2020-05-14 17:35:45

Fortunately, our work so far is most of the proof of the second part of the conjecture: why can't we include any numbers with a $2$ as a digit in the sequence?

Fortunately, our work so far is most of the proof of the second part of the conjecture: why can't we include any numbers with a $2$ as a digit in the sequence?

rrusczyk
2020-05-14 17:35:48

Let's try an example. Suppose we tried to insert $z = 120221_3$ into our sequence as we're building it. Remember, we've already included all the smaller numbers in the sequence that consist only of $0$s and $1$s in base $3.$

Let's try an example. Suppose we tried to insert $z = 120221_3$ into our sequence as we're building it. Remember, we've already included all the smaller numbers in the sequence that consist only of $0$s and $1$s in base $3.$

rrusczyk
2020-05-14 17:36:06

Can you find an $x$ and a $y$ already in the sequence -- so that $x$ and $y$ only has $0$s and $1$s -- so that $x,y,z$ are an arithmetic sequence?

Can you find an $x$ and a $y$ already in the sequence -- so that $x$ and $y$ only has $0$s and $1$s -- so that $x,y,z$ are an arithmetic sequence?

EFrame2
2020-05-14 17:37:34

100001 110111

100001 110111

edjar
2020-05-14 17:37:41

x=100001 y= 110111

x=100001 y= 110111

huela
2020-05-14 17:37:42

100001, 110111, 120221

100001, 110111, 120221

rrusczyk
2020-05-14 17:37:52

Just take $x$ to be the same number as $z$, but with the $2$s replaced by $0$s.

Just take $x$ to be the same number as $z$, but with the $2$s replaced by $0$s.

rrusczyk
2020-05-14 17:37:55

So $x = 100001_3.$

So $x = 100001_3.$

rrusczyk
2020-05-14 17:37:59

And what happens when we add $x+z?$

And what happens when we add $x+z?$

sixoneeight
2020-05-14 17:38:25

you get only 2s

you get only 2s

MathBluebird
2020-05-14 17:38:25

we get 2y

we get 2y

sixoneeight
2020-05-14 17:38:25

2s and 0s only.

2s and 0s only.

Puffer13
2020-05-14 17:38:25

220222

220222

huela
2020-05-14 17:38:25

220222

220222

AGENT99
2020-05-14 17:38:25

220222

220222

kdraganov
2020-05-14 17:38:25

$220222_3$

$220222_3$

fasterthanlight
2020-05-14 17:38:27

220222

220222

Lux1
2020-05-14 17:38:28

the y has only 2s and 0s

the y has only 2s and 0s

rrusczyk
2020-05-14 17:38:33

Again, there's no carrying! If $z$ has a $0$ or $1,$ then $x$ does too, but if $z$ has a $2,$ then $x$ has a $0.$

Again, there's no carrying! If $z$ has a $0$ or $1,$ then $x$ does too, but if $z$ has a $2,$ then $x$ has a $0.$

rrusczyk
2020-05-14 17:38:34

And furthermore, all the digit sums are $0$s or $2$s! In our example $x+z = 220222_3.$

And furthermore, all the digit sums are $0$s or $2$s! In our example $x+z = 220222_3.$

rrusczyk
2020-05-14 17:38:35

That means we can divide by $2!$

That means we can divide by $2!$

rrusczyk
2020-05-14 17:38:37

All the $2$s become $1$s, and we have $y = 110111_3$ with all $0$s and $1$s.

All the $2$s become $1$s, and we have $y = 110111_3$ with all $0$s and $1$s.

rrusczyk
2020-05-14 17:39:25

This process generalizes! Given any $z$, take $x$ to be the same as $z$ but with its $2$s replaced by $0$s. Then $x+z$ will have only $0$s and $2$s, so $y = \frac12(x+z)$ will have only $0$s and $1$s, and $x,y,z$ is an arithmetic sequence. (Notice that $y$ is what you get when you replace the $2$s in $z$s with $1$s.)

This process generalizes! Given any $z$, take $x$ to be the same as $z$ but with its $2$s replaced by $0$s. Then $x+z$ will have only $0$s and $2$s, so $y = \frac12(x+z)$ will have only $0$s and $1$s, and $x,y,z$ is an arithmetic sequence. (Notice that $y$ is what you get when you replace the $2$s in $z$s with $1$s.)

Puffer13
2020-05-14 17:39:36

yaaah

yaaah

RedPenMan
2020-05-14 17:39:36

woah

woah

cj13609517288
2020-05-14 17:39:36

Wait, but that's it!

Wait, but that's it!

sdattilo2002
2020-05-14 17:39:36

cool!

cool!

edjar
2020-05-14 17:39:40

We can do this with any number so our proof is complete and the answer to the problem is 64

We can do this with any number so our proof is complete and the answer to the problem is 64

EulerRocks2.718
2020-05-14 17:39:47

Cool!

Cool!

MathIsFun286
2020-05-14 17:39:47

wow

wow

Math5K
2020-05-14 17:39:47

wow!

wow!

danprathab
2020-05-14 17:39:49

AND WE ARE DONE WITH OUR PROOF ! ! !

AND WE ARE DONE WITH OUR PROOF ! ! !

waddlethegreat
2020-05-14 17:39:51

which proves that 64 or 2^6 is the answer!

which proves that 64 or 2^6 is the answer!

Lux1
2020-05-14 17:39:55

wow. this problem is neat!

wow. this problem is neat!

CHIPPER33
2020-05-14 17:39:55

That's amazing!

That's amazing!

PShucks
2020-05-14 17:39:55

yay

yay

missionsqhc
2020-05-14 17:39:55

nice

nice

PShucks
2020-05-14 17:39:55

wow

wow

tigerjade003
2020-05-14 17:39:55

yay

yay

NinjaMango
2020-05-14 17:40:01

Wow!

Wow!

jupiter314
2020-05-14 17:40:01

great!

great!

aidni47
2020-05-14 17:40:01

wow!

wow!

mfro24
2020-05-14 17:40:01

That was awesome!

That was awesome!

JBB
2020-05-14 17:40:01

wow

wow

donguri
2020-05-14 17:40:01

YAY!

YAY!

akpi
2020-05-14 17:40:01

yay

yay

Tiger2010
2020-05-14 17:40:01

brilliant

brilliant

nathanqiu
2020-05-14 17:40:04

AND WE'RE DONE!!!

AND WE'RE DONE!!!

doglover07
2020-05-14 17:40:04

yay

yay

Chesssaga
2020-05-14 17:40:04

YAYAYAYAYAYAY

YAYAYAYAYAYAY

rrusczyk
2020-05-14 17:40:08

So we've proved the conjecture, and now proved that the answer is indeed $\boxed{64}$.

So we've proved the conjecture, and now proved that the answer is indeed $\boxed{64}$.

rrusczyk
2020-05-14 17:40:17

Yeah, that proof was pretty tough!

Yeah, that proof was pretty tough!

rrusczyk
2020-05-14 17:40:33

Totally cool if you didn't follow it -- we went through it a lot faster than it took me to find it

Totally cool if you didn't follow it -- we went through it a lot faster than it took me to find it

Jomo
2020-05-14 17:40:43

Are we going to do the target problems?

Are we going to do the target problems?

donguri
2020-05-14 17:40:43

Are we doing any target problems?

Are we doing any target problems?

sosiaops
2020-05-14 17:40:46

are we doing target round?

are we doing target round?

rrusczyk
2020-05-14 17:40:56

If by "we" you mean "DPatrick", why, yes!

If by "we" you mean "DPatrick", why, yes!

DPatrick
2020-05-14 17:41:10

OK, now you can get your calculator out. We're switching to Target Round problems!

OK, now you can get your calculator out. We're switching to Target Round problems!

onedance
2020-05-14 17:41:29

dun dun dun!

dun dun dun!

Kush0924
2020-05-14 17:41:29

Target is the best

Target is the best

evanshawn316
2020-05-14 17:41:29

AWESOME!

AWESOME!

DPatrick
2020-05-14 17:41:37

**Target #6:**Iris is playing a game that has a $5 \times 5$ gameboard like the one shown. The goal is to get her game piece from the square labeled $\star$ to the square labeled $\circ$ using a series of moves any positive integer number of squares up or any positive integer number of squares to the right. Note that moving two squares up in a single move is different than moving two squares up in two moves. How many unique sequences of moves can Iris make to get her game piece from $\star$ to $\circ$?
DPatrick
2020-05-14 17:41:43

DPatrick
2020-05-14 17:42:09

Any suggestions?

Any suggestions?

Leonard_my_dude
2020-05-14 17:42:50

count the number of ways to each square

count the number of ways to each square

yayy
2020-05-14 17:42:50

count the ways to get to every square

count the ways to get to every square

Asterlan
2020-05-14 17:42:50

write the number of ways to get to each square in them

write the number of ways to get to each square in them

Makorn
2020-05-14 17:43:45

start filling squares in with how many ways u can get from the starred square to that square; u can calculate this by summing the ways from all possible previous squares

start filling squares in with how many ways u can get from the starred square to that square; u can calculate this by summing the ways from all possible previous squares

austinchen2005
2020-05-14 17:43:45

you can fill in squares with number of ways, and you figure out how many number of ways on a square from the sum of the squares to the left and below it

you can fill in squares with number of ways, and you figure out how many number of ways on a square from the sum of the squares to the left and below it

bishope6
2020-05-14 17:43:59

Start by figuring out how many sequences of moves there are to get to the star from the squares nearest to the star, and work backwards.

Start by figuring out how many sequences of moves there are to get to the star from the squares nearest to the star, and work backwards.

DPatrick
2020-05-14 17:44:26

I think this is a good general strategy: we'll try to count the number of paths to or from each square.

I think this is a good general strategy: we'll try to count the number of paths to or from each square.

DPatrick
2020-05-14 17:44:47

It doesn't really matter too much whether we start at the star and count forward, or start at the circle and count backwards.

It doesn't really matter too much whether we start at the star and count forward, or start at the circle and count backwards.

DPatrick
2020-05-14 17:45:25

But because I wrote my diagrams for the second way, let's work backwards. (It's pretty much the same count.)

But because I wrote my diagrams for the second way, let's work backwards. (It's pretty much the same count.)

DPatrick
2020-05-14 17:45:38

That is, we start at the $\circ$ and count backwards how many ways to get to $\circ$ for each space.

That is, we start at the $\circ$ and count backwards how many ways to get to $\circ$ for each space.

MathWizard38025
2020-05-14 17:45:50

but wont that take a really long time

but wont that take a really long time

DPatrick
2020-05-14 17:46:00

I hope not! The grid isn't too big. Let's see what happens.

I hope not! The grid isn't too big. Let's see what happens.

DPatrick
2020-05-14 17:46:11

We start at $\circ$. There's only $1$ way to get to $\circ$ from $\circ$--you're already there!

We start at $\circ$. There's only $1$ way to get to $\circ$ from $\circ$--you're already there!

DPatrick
2020-05-14 17:46:15

What about the two spaces immediately adjacent to $\circ$?

What about the two spaces immediately adjacent to $\circ$?

rjiangbz
2020-05-14 17:46:31

1

1

Math5K
2020-05-14 17:46:31

1

1

aidni47
2020-05-14 17:46:31

1

1

Superior
2020-05-14 17:46:31

1 way

1 way

vsurya
2020-05-14 17:46:31

1

1

SkywalkerAUV
2020-05-14 17:46:31

each 1

each 1

jbear911
2020-05-14 17:46:31

1

1

egerns
2020-05-14 17:46:31

1

1

EulerRocks2.718
2020-05-14 17:46:31

1 way

1 way

green_alligator
2020-05-14 17:46:31

1 way

1 way

bobthegod78
2020-05-14 17:46:31

1 way

1 way

MathIsFun286
2020-05-14 17:46:31

1 and 1

1 and 1

Maths4J
2020-05-14 17:46:31

both 1

both 1

DPatrick
2020-05-14 17:46:35

There's also only $1$ way to get to $\circ$ from each of the spaces adjacent to $\circ$.

There's also only $1$ way to get to $\circ$ from each of the spaces adjacent to $\circ$.

DPatrick
2020-05-14 17:46:49

Let's keep track with a picture.

Let's keep track with a picture.

DPatrick
2020-05-14 17:46:55

DPatrick
2020-05-14 17:47:04

Next, let's consider the square with the red $X.$ How many ways to get to $\circ$ from $X?$

Next, let's consider the square with the red $X.$ How many ways to get to $\circ$ from $X?$

Ninja11
2020-05-14 17:47:22

x=2

x=2

nathanqiu
2020-05-14 17:47:22

X=2

X=2

winterrain01
2020-05-14 17:47:22

2

2

MathAMC8
2020-05-14 17:47:22

2 way

2 way

Eng123
2020-05-14 17:47:22

2

2

scoutskylar
2020-05-14 17:47:22

2

2

JBB
2020-05-14 17:47:22

2

2

eez
2020-05-14 17:47:22

2

2

Ninja11
2020-05-14 17:47:22

2

2

Mathemats
2020-05-14 17:47:22

2

2

SharonW
2020-05-14 17:47:22

2

2

yukrant1
2020-05-14 17:47:22

2

2

huela
2020-05-14 17:47:22

2

2

yayatheduck
2020-05-14 17:47:22

2

2

Sri_Math
2020-05-14 17:47:22

x=2

x=2

superagh
2020-05-14 17:47:22

2

2

DPatrick
2020-05-14 17:47:32

From there, we can take one step right (and then there is $1$ way to finish).

Or we can take one step up (and then there is $1$ way to finish).

From there, we can take one step right (and then there is $1$ way to finish).

Or we can take one step up (and then there is $1$ way to finish).

DPatrick
2020-05-14 17:47:37

Combining these, we have $1+1=2$ ways to finish from $X.$

Combining these, we have $1+1=2$ ways to finish from $X.$

DPatrick
2020-05-14 17:47:44

DPatrick
2020-05-14 17:47:54

What about $Y$? How many ways to $\circ$ from $Y?$

What about $Y$? How many ways to $\circ$ from $Y?$

ChrisalonaLiverspur
2020-05-14 17:48:08

$2$ ways.

$2$ ways.

Floating-Clouds
2020-05-14 17:48:08

2 ways.

2 ways.

akpi2
2020-05-14 17:48:08

2

2

PShucks
2020-05-14 17:48:08

2

2

mathgenius237
2020-05-14 17:48:08

2

2

MathWiz20
2020-05-14 17:48:08

2

2

PShucks
2020-05-14 17:48:08

2!

2!

Poki
2020-05-14 17:48:08

2

2

PShucks
2020-05-14 17:48:08

2

2

m_goli
2020-05-14 17:48:08

2

2

albertedwin
2020-05-14 17:48:08

2

2

DPatrick
2020-05-14 17:48:15

We can take one step right, with $1$ way to finish from there.

Or, we can take two steps right, with $1$ way to finish from there--already done!

We can take one step right, with $1$ way to finish from there.

Or, we can take two steps right, with $1$ way to finish from there--already done!

DPatrick
2020-05-14 17:48:21

Combining these, we have $1+1=2$ ways to finish from $Y.$

Combining these, we have $1+1=2$ ways to finish from $Y.$

bobthegod78
2020-05-14 17:48:38

same for y and z

same for y and z

jupiter314
2020-05-14 17:48:38

y and z are also equal to 2

y and z are also equal to 2

nathanqiu
2020-05-14 17:48:38

$Y,Z=2$

$Y,Z=2$

mfro24
2020-05-14 17:48:38

$2$ ways to $Z$ as well!

$2$ ways to $Z$ as well!

tigerjade003
2020-05-14 17:48:38

Z = 2 as well

Z = 2 as well

Kruxe
2020-05-14 17:48:38

so there's 2 ways from Z as well

so there's 2 ways from Z as well

nmadhu
2020-05-14 17:48:38

2

2

tumbleweed
2020-05-14 17:48:38

2

2

ARSM2019
2020-05-14 17:48:43

same thing for z right

same thing for z right

DPatrick
2020-05-14 17:48:48

The situation at $Z$ is identical to that at $Y$, except that we are going up instead of to the right. We'll keep exploiting this symmetry throughout this solution!

The situation at $Z$ is identical to that at $Y$, except that we are going up instead of to the right. We'll keep exploiting this symmetry throughout this solution!

DPatrick
2020-05-14 17:48:52

Let's update the diagram and label the next group of squares:

Let's update the diagram and label the next group of squares:

DPatrick
2020-05-14 17:48:57

DPatrick
2020-05-14 17:49:11

How many ways to finish from $A?$

How many ways to finish from $A?$

Leonard_my_dude
2020-05-14 17:49:38

4

4

CrazyVideoGamez
2020-05-14 17:49:38

4

4

gs_2006
2020-05-14 17:49:38

4

4

ellnoo
2020-05-14 17:49:38

4

4

walrus987
2020-05-14 17:49:38

4

4

Jalenluorion
2020-05-14 17:49:38

4

4

nikenissan
2020-05-14 17:49:38

4 ways

4 ways

JQWERTY6
2020-05-14 17:49:38

4

4

lihao_david
2020-05-14 17:49:38

4

4

RedFireTruck
2020-05-14 17:49:38

4

4

mfro24
2020-05-14 17:49:38

$A=4$

$A=4$

sdattilo2002
2020-05-14 17:49:38

4

4

sigma_notation
2020-05-14 17:49:38

4

4

aops_band
2020-05-14 17:49:38

4

4

Pianodude
2020-05-14 17:49:38

4

4

DPatrick
2020-05-14 17:49:43

We can jump from $A$ to any of the three squares to the right of $A.$

We can jump from $A$ to any of the three squares to the right of $A.$

DPatrick
2020-05-14 17:49:55

And from those squares, we have $2$, $1$, and $1$ way to finish.

And from those squares, we have $2$, $1$, and $1$ way to finish.

DPatrick
2020-05-14 17:50:00

So we have $2+1+1 = 4$ ways to get from $A$ to $\circ$.

So we have $2+1+1 = 4$ ways to get from $A$ to $\circ$.

mfro24
2020-05-14 17:50:13

$D=4$

$D=4$

sixoneeight
2020-05-14 17:50:13

D is 4 too!

D is 4 too!

jbear911
2020-05-14 17:50:13

same with D

same with D

ALDW123
2020-05-14 17:50:13

d=4 as well

d=4 as well

Chesssaga
2020-05-14 17:50:13

Same with D

Same with D

MathIsFun286
2020-05-14 17:50:13

$d=4$

$d=4$

winterrain01
2020-05-14 17:50:13

By symmetry, D is 4 too, right?

By symmetry, D is 4 too, right?

DPatrick
2020-05-14 17:50:26

Yep: by the symmetry of the grid, we have $4$ ways from $D$ too.

Yep: by the symmetry of the grid, we have $4$ ways from $D$ too.

DPatrick
2020-05-14 17:50:30

And what about $B$?

And what about $B$?

Awesome3.14
2020-05-14 17:51:06

B=5 ways

B=5 ways

TheEpicCarrot7
2020-05-14 17:51:06

B=5

B=5

Streaks123
2020-05-14 17:51:06

5

5

XTJin
2020-05-14 17:51:06

5

5

Awesome3.14
2020-05-14 17:51:06

5

5

lrjr24
2020-05-14 17:51:06

5.

5.

Math4Life2020
2020-05-14 17:51:06

B = 5

B = 5

john0512
2020-05-14 17:51:06

5

5

CHIPPER33
2020-05-14 17:51:06

b=5

b=5

sarahAops2020
2020-05-14 17:51:06

5

5

mfro24
2020-05-14 17:51:06

$B=5$

$B=5$

MR_67
2020-05-14 17:51:06

B and C = 5

B and C = 5

CT17
2020-05-14 17:51:06

5

5

MrEgggga
2020-05-14 17:51:06

$B=5$

$B=5$

awesomeming327.
2020-05-14 17:51:06

5.

5.

MR_67
2020-05-14 17:51:06

B=5

B=5

yayy
2020-05-14 17:51:06

5

5

Kason
2020-05-14 17:51:06

B is 5, 3+2, and C is by symmetry, 5,too

B is 5, 3+2, and C is by symmetry, 5,too

DPatrick
2020-05-14 17:51:10

From $B,$ we can go one step up ($2$ ways to finish), one step right ($2$ ways to finish), or two steps right ($1$ way to finish).

From $B,$ we can go one step up ($2$ ways to finish), one step right ($2$ ways to finish), or two steps right ($1$ way to finish).

DPatrick
2020-05-14 17:51:18

So there are $2+2+1=5$ ways to finish from $B.$

So there are $2+2+1=5$ ways to finish from $B.$

DPatrick
2020-05-14 17:51:26

And the same for $C$, by symmetry.

And the same for $C$, by symmetry.

DPatrick
2020-05-14 17:51:31

DPatrick
2020-05-14 17:51:39

Now what?

Now what?

Math5K
2020-05-14 17:52:00

continue

continue

MathIsFun286
2020-05-14 17:52:00

next diagonal

next diagonal

sixoneeight
2020-05-14 17:52:00

we keep going

we keep going

CrazyVideoGamez
2020-05-14 17:52:00

Keep going on to the next squares!

Keep going on to the next squares!

nathanqiu
2020-05-14 17:52:03

this remeinds me of pascal's triangle

this remeinds me of pascal's triangle

Jalenluorion
2020-05-14 17:52:20

looks a little like pascals riangle

looks a little like pascals riangle

sdattilo2002
2020-05-14 17:52:20

is this pascal's triangle in a weird way

is this pascal's triangle in a weird way

DPatrick
2020-05-14 17:52:28

Right...it's a bit like Pascal's Triangle.

Right...it's a bit like Pascal's Triangle.

DPatrick
2020-05-14 17:52:59

Except that, instead of adding just the two adjacent numbers, we're adding

Except that, instead of adding just the two adjacent numbers, we're adding

**all**the numbers in the squares directly to the right of that square and**all**the numbers directly above that square.
walrus987
2020-05-14 17:53:07

it's pascal's except each thing is the sum of everything to the right and above, rather than the immediate right and above

it's pascal's except each thing is the sum of everything to the right and above, rather than the immediate right and above

DPatrick
2020-05-14 17:53:21

So let's work on the next diagonal:

So let's work on the next diagonal:

DPatrick
2020-05-14 17:53:25

DPatrick
2020-05-14 17:53:33

Note that at this point I've built the symmetry into my labels: the two $E$s will be the same, as will the two $F$s.

Note that at this point I've built the symmetry into my labels: the two $E$s will be the same, as will the two $F$s.

ARSM2019
2020-05-14 17:53:55

all the way on the left is 8 then because 4+2+1+1=8

all the way on the left is 8 then because 4+2+1+1=8

Kruxe
2020-05-14 17:53:55

so that means the top-left corner is 8?

so that means the top-left corner is 8?

Pianodude
2020-05-14 17:53:55

E=8

E=8

PShucks
2020-05-14 17:53:55

e=8

e=8

ARSM2019
2020-05-14 17:53:55

E=8

E=8

MR_67
2020-05-14 17:53:55

E = 8

E = 8

menlo
2020-05-14 17:53:55

E=8

E=8

sri1priya
2020-05-14 17:53:55

e = 8

e = 8

anishcool11
2020-05-14 17:53:55

E is 8

E is 8

DarthMaul
2020-05-14 17:53:55

E: 8

E: 8

Doudou_Chen
2020-05-14 17:53:55

E has 8

E has 8

DPatrick
2020-05-14 17:54:00

From the top $E$, we sum the numbers to the right of $E$ and get $4+2+1+1 = 8.$

From the top $E$, we sum the numbers to the right of $E$ and get $4+2+1+1 = 8.$

TheMagician
2020-05-14 17:54:26

f = 12

f = 12

ccu700037
2020-05-14 17:54:26

F=12

F=12

mathbw225
2020-05-14 17:54:26

F=12

F=12

bobjoebilly
2020-05-14 17:54:26

F = 4 + 5 + 2 + 1 = 12

F = 4 + 5 + 2 + 1 = 12

Dalar25
2020-05-14 17:54:26

F=12

F=12

Juneybug
2020-05-14 17:54:26

f=12

f=12

bronzetruck2016
2020-05-14 17:54:26

f=12

f=12

akpi2
2020-05-14 17:54:26

f=12

f=12

wertiol123
2020-05-14 17:54:26

F=12

F=12

mfro24
2020-05-14 17:54:26

$F = [4]+[1+2+5] = 12$

$F = [4]+[1+2+5] = 12$

DPatrick
2020-05-14 17:54:30

From the top $F$, we sum the numbers above and to the right of $F$ and get $4+5+2+1 = 12.$

From the top $F$, we sum the numbers above and to the right of $F$ and get $4+5+2+1 = 12.$

Math5K
2020-05-14 17:54:51

G=14

G=14

Dalar25
2020-05-14 17:54:51

G=14

G=14

onionheadjr
2020-05-14 17:54:51

G=14

G=14

mohanty
2020-05-14 17:54:51

G is 14

G is 14

jjw4106
2020-05-14 17:54:51

G=14

G=14

blue_arrow
2020-05-14 17:54:51

G=14

G=14

aidni47
2020-05-14 17:54:51

g=14

g=14

EpicGirl
2020-05-14 17:54:51

G=14

G=14

MathBluebird
2020-05-14 17:54:51

g = 14

g = 14

Math5K
2020-05-14 17:54:51

G=14.

G=14.

Flying_Bird
2020-05-14 17:54:51

$G=14$

$G=14$

DPatrick
2020-05-14 17:54:55

From $G$, we sum the numbers above and to the right of $G$ and get $2+5+5+2 = 14.$

From $G$, we sum the numbers above and to the right of $G$ and get $2+5+5+2 = 14.$

DPatrick
2020-05-14 17:54:59

DPatrick
2020-05-14 17:55:04

We keep going like this!

We keep going like this!

DPatrick
2020-05-14 17:55:07

What numbers are on the next diagonal?

What numbers are on the next diagonal?

albertedwin
2020-05-14 17:55:42

28

28

evanshawn316
2020-05-14 17:55:42

28

28

vcric247
2020-05-14 17:55:42

28

28

bobthegod78
2020-05-14 17:55:42

28, 37

28, 37

Ninja11
2020-05-14 17:55:42

37

37

Lux1
2020-05-14 17:55:42

first one is 28

first one is 28

asbodke
2020-05-14 17:55:42

28 and 37

28 and 37

Maths4J
2020-05-14 17:55:42

28 and 37

28 and 37

fasterthanlight
2020-05-14 17:55:42

28,37,37,28

28,37,37,28

bot101
2020-05-14 17:55:42

28 and 37

28 and 37

vsurya
2020-05-14 17:55:42

28

28

AmpharosX
2020-05-14 17:55:42

28,37,37,28

28,37,37,28

evanshawn316
2020-05-14 17:55:42

28 and 37

28 and 37

pvmathnerd
2020-05-14 17:55:42

28 and 37

28 and 37

DPatrick
2020-05-14 17:55:46

The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$

The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$

DPatrick
2020-05-14 17:55:52

DPatrick
2020-05-14 17:55:54

And next?

And next?

sigma_notation
2020-05-14 17:56:27

94,106,94

94,106,94

Ninja-Girl
2020-05-14 17:56:27

94 106 94

94 106 94

mohanty
2020-05-14 17:56:27

94 and 106

94 and 106

Awesome3.14
2020-05-14 17:56:27

94 and 106

94 and 106

lihao_david
2020-05-14 17:56:27

94 and 106

94 and 106

Ninja11
2020-05-14 17:56:27

94 and 106

94 and 106

jupiter314
2020-05-14 17:56:27

94 and 106

94 and 106

Mathematicalbrain
2020-05-14 17:56:27

94 and 106

94 and 106

DPatrick
2020-05-14 17:56:48

Oops, I got out of sync by one diagonal!

Oops, I got out of sync by one diagonal!

DPatrick
2020-05-14 17:56:57

The prior diagonal was $8+12+5+2+1 = 28$ and $4+12+15+5+2 = 37.$

The prior diagonal was $8+12+5+2+1 = 28$ and $4+12+15+5+2 = 37.$

DPatrick
2020-05-14 17:57:04

The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$

The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$

DPatrick
2020-05-14 17:57:08

DPatrick
2020-05-14 17:57:20

Getting near the finish line!

Getting near the finish line!

FearlessTaurus
2020-05-14 17:57:38

289 is next

289 is next

bobjoebilly
2020-05-14 17:57:38

then 289

then 289

evanshawn316
2020-05-14 17:57:38

289

289

MR_67
2020-05-14 17:57:38

Next is 289 and 289

Next is 289 and 289

cat_maniac_8
2020-05-14 17:57:38

289 is the last ones!!!

289 is the last ones!!!

Ninja11
2020-05-14 17:57:38

289

289

gordonhero
2020-05-14 17:57:38

289

289

yukrant1
2020-05-14 17:57:38

289,289 is next diagonal

289,289 is next diagonal

TheEpicCarrot7
2020-05-14 17:57:38

289

289

DPatrick
2020-05-14 17:57:42

The next diagonal has $8+28+94+106+37+12+4 = 289.$

The next diagonal has $8+28+94+106+37+12+4 = 289.$

DPatrick
2020-05-14 17:57:56

DPatrick
2020-05-14 17:58:08

And the lower-left square, which is the final answer?

And the lower-left square, which is the final answer?

donguri
2020-05-14 17:58:28

The answer is $\boxed{838}$

The answer is $\boxed{838}$

CHIPPER33
2020-05-14 17:58:28

838

838

yayatheduck
2020-05-14 17:58:28

and finally 838!

and finally 838!

CHIPPER33
2020-05-14 17:58:28

838!!!

838!!!

Liakas_42
2020-05-14 17:58:28

838

838

Quaoar
2020-05-14 17:58:28

838

838

winterrain01
2020-05-14 17:58:28

838 is the answer!!!

838 is the answer!!!

lucaszhou
2020-05-14 17:58:28

838

838

nathanqiu
2020-05-14 17:58:28

838!!!

838!!!

Sedro
2020-05-14 17:58:28

838

838

scoutskylar
2020-05-14 17:58:28

$\bbox[5px, border: 2px solid black]{838}$

$\bbox[5px, border: 2px solid black]{838}$

concierge
2020-05-14 17:58:28

838

838

Zhaom
2020-05-14 17:58:28

838

838

tigerjade003
2020-05-14 17:58:28

so finnally, it is 838

so finnally, it is 838

DPatrick
2020-05-14 17:58:33

The final answer is that there are $8+28+94+289+289+94+28+8 = \boxed{838}$ paths from $\star$ to $\circ$.

The final answer is that there are $8+28+94+289+289+94+28+8 = \boxed{838}$ paths from $\star$ to $\circ$.

DPatrick
2020-05-14 17:58:38

DPatrick
2020-05-14 17:59:06

Sometimes just getting your handy dirty and computing is the simplest way to the answer.

Sometimes just getting your handy dirty and computing is the simplest way to the answer.

green_alligator
2020-05-14 17:59:26

is there a faster way to do it?

is there a faster way to do it?

gs_2006
2020-05-14 17:59:26

is there an easier way or is this the only one

is there an easier way or is this the only one

pianoboy
2020-05-14 17:59:26

that was long. How could I do that in a shorter way

that was long. How could I do that in a shorter way

DPatrick
2020-05-14 17:59:51

Not that I'm readily aware of. I don't really recognize the numbers we're seeing, so there's no obvious pattern that jumps out at me.

Not that I'm readily aware of. I don't really recognize the numbers we're seeing, so there's no obvious pattern that jumps out at me.

DPatrick
2020-05-14 18:00:12

OK, back to rrusczyk for some watermelon.

OK, back to rrusczyk for some watermelon.

rrusczyk
2020-05-14 18:00:19

Yum!

Yum!

rrusczyk
2020-05-14 18:00:22

**Target #7:**The figure shows a perfectly round watermelon, represented by circle $O,$ secured on a flat table, represented by segment $AB,$ using a rope. One end of the rope is attached to the table at $A,$ then draped over the top of the watermelon, and the other end is attached at $B.$ The watermelon has diameter $20$ cm. The center of the watermelon is $20$ cm from point $A$ and $10\sqrt{2}$ cm from point $B.$ What is the minimum length of rope necessary to secure the watermelon to the table? Express your answer as a decimal to the nearest tenth.
rrusczyk
2020-05-14 18:00:33

rrusczyk
2020-05-14 18:00:36

Just so we're clear on the problem: what we're trying to do is to measure the blue length shown:

Just so we're clear on the problem: what we're trying to do is to measure the blue length shown:

rrusczyk
2020-05-14 18:00:40

rrusczyk
2020-05-14 18:00:51

Where should we start?

Where should we start?

nmadhu
2020-05-14 18:01:23

draw radii to tangents

draw radii to tangents

Jalenluorion
2020-05-14 18:01:23

draw a radius from o to ab

draw a radius from o to ab

MathIsFun286
2020-05-14 18:01:23

radii to the points of tangency

radii to the points of tangency

nmadhu
2020-05-14 18:01:23

draw radii to tangent points

draw radii to tangent points

SuperSine
2020-05-14 18:01:23

draw lines to the points of tangency

draw lines to the points of tangency

yayy
2020-05-14 18:01:27

draw the radii perpendicular to tangents

draw the radii perpendicular to tangents

memi
2020-05-14 18:01:31

drawing radii to points of tangency

drawing radii to points of tangency

rrusczyk
2020-05-14 18:01:37

It's usually help to draw in radii for a circle where we think it will help.

It's usually help to draw in radii for a circle where we think it will help.

rrusczyk
2020-05-14 18:01:42

Here, there are three radii just begging to be drawn -- radii to points of tangency! That will give us right angles, and we like right angles.

Here, there are three radii just begging to be drawn -- radii to points of tangency! That will give us right angles, and we like right angles.

rrusczyk
2020-05-14 18:01:45

Let's draw the radii to the two points where the rope starts and stops running along the melon (and we'll call those points $C$ and $D$).

Let's draw the radii to the two points where the rope starts and stops running along the melon (and we'll call those points $C$ and $D$).

rrusczyk
2020-05-14 18:01:46

Let's also draw the radius to where the melon touches the table (and we'll call that point $E$).

Let's also draw the radius to where the melon touches the table (and we'll call that point $E$).

rrusczyk
2020-05-14 18:01:54

And let's add the lengths we know to the picture. All the radii are $10$, and we have the two given lengths $OA = 20$ and $OB = 10\sqrt2.$

And let's add the lengths we know to the picture. All the radii are $10$, and we have the two given lengths $OA = 20$ and $OB = 10\sqrt2.$

rrusczyk
2020-05-14 18:01:57

rrusczyk
2020-05-14 18:02:09

So our goal is to compute the total length $AC + \widehat{CD} + DB$, where $\widehat{CD}$ means the arc length from $C$ to $D.$

So our goal is to compute the total length $AC + \widehat{CD} + DB$, where $\widehat{CD}$ means the arc length from $C$ to $D.$

rrusczyk
2020-05-14 18:02:14

What's $AC?$

What's $AC?$

KingRavi
2020-05-14 18:02:42

AOC is 30 60 90 triangle

AOC is 30 60 90 triangle

KingRavi
2020-05-14 18:02:42

10 sqrt 3

10 sqrt 3

Derpy123
2020-05-14 18:02:42

10sqrt3

10sqrt3

xMidnightFirex
2020-05-14 18:02:42

10sqrt3

10sqrt3

SkywalkerAUV
2020-05-14 18:02:42

10sqrt3

10sqrt3

Jalenluorion
2020-05-14 18:02:42

special right triangles!

special right triangles!

harmonyguan
2020-05-14 18:02:42

10sqrt3

10sqrt3

lamphead
2020-05-14 18:02:42

$10sqrt{3}$

$10sqrt{3}$

Dalar25
2020-05-14 18:02:42

10*sqrt(3)

10*sqrt(3)

ARSM2019
2020-05-14 18:02:42

ac=10 root 3

ac=10 root 3

Leonard_my_dude
2020-05-14 18:02:42

10sqrt3

10sqrt3

Speedstorm
2020-05-14 18:02:42

$10\sqrt{3}$

$10\sqrt{3}$

Maths4J
2020-05-14 18:02:42

AC=10sqrt(3)

AC=10sqrt(3)

winner320
2020-05-14 18:02:42

10 root 3

10 root 3

LightiningBlazer
2020-05-14 18:02:42

10sqrt(3)

10sqrt(3)

Math5K
2020-05-14 18:02:42

10sqrt(3)

10sqrt(3)

asbodke
2020-05-14 18:02:42

$10\sqrt{3}$

$10\sqrt{3}$

Pianodude
2020-05-14 18:02:42

AC = 10sqrt3

AC = 10sqrt3

Sri_Math
2020-05-14 18:02:42

10√3

10√3

aidni47
2020-05-14 18:02:42

10 rt 3

10 rt 3

lamphead
2020-05-14 18:02:42

$10\sqrt{3}$

$10\sqrt{3}$

tumbleweed
2020-05-14 18:02:42

10\sqrt3

10\sqrt3

rrusczyk
2020-05-14 18:02:54

Side $OC$ is half of side $OA.$

Side $OC$ is half of side $OA.$

rrusczyk
2020-05-14 18:02:57

So, $AOC$ is a 30-60-90 triangle!

So, $AOC$ is a 30-60-90 triangle!

rrusczyk
2020-05-14 18:02:59

So $AC = 10\sqrt3$.

So $AC = 10\sqrt3$.

rrusczyk
2020-05-14 18:03:00

How about $DB$?

How about $DB$?

xMidnightFirex
2020-05-14 18:03:48

we have square ODBE

we have square ODBE

Math5K
2020-05-14 18:03:48

30-60-90 and 45-45-90 triangles!

30-60-90 and 45-45-90 triangles!

scoutskylar
2020-05-14 18:03:48

$10$

$10$

krishgarg
2020-05-14 18:03:48

10

10

CHIPPER33
2020-05-14 18:03:48

10, 45-45-90

10, 45-45-90

vrondoS
2020-05-14 18:03:48

10

10

Significant
2020-05-14 18:03:48

10

10

sosiaops
2020-05-14 18:03:48

$10$

$10$

tientien1
2020-05-14 18:03:48

10

10

Jerry_Guo
2020-05-14 18:03:48

10

10

JunoSuno
2020-05-14 18:03:48

10

10

millburn2006
2020-05-14 18:03:48

10

10

punkinpiday
2020-05-14 18:03:48

DB = 10

DB = 10

akpi
2020-05-14 18:03:48

10

10

scinderella220
2020-05-14 18:03:48

10

10

Scifan671
2020-05-14 18:03:48

10

10

akpi
2020-05-14 18:03:48

it is 10

it is 10

sosiaops
2020-05-14 18:03:48

$10$ !

$10$ !

Doudou_Chen
2020-05-14 18:03:48

45-45-90 triangle, so DB = 10!

45-45-90 triangle, so DB = 10!

Lux1
2020-05-14 18:03:48

it has a length of 10

it has a length of 10

KSH31415
2020-05-14 18:03:48

10 because it is a square

10 because it is a square

rrusczyk
2020-05-14 18:03:51

$ODB$ is a 45-45-90 triangle! (Or you might have seen that $ODBE$ is a square.)

$ODB$ is a 45-45-90 triangle! (Or you might have seen that $ODBE$ is a square.)

rrusczyk
2020-05-14 18:03:52

So $DB = 10$.

So $DB = 10$.

rrusczyk
2020-05-14 18:03:53

And what about the arc $\widehat{CD}$?

And what about the arc $\widehat{CD}$?

Ninja11
2020-05-14 18:04:16

150 degress

150 degress

Pokemon2
2020-05-14 18:04:16

150 degree measure?

150 degree measure?

harmonyguan
2020-05-14 18:04:16

150 degrees

150 degrees

Liakas_42
2020-05-14 18:04:16

150

150

winner320
2020-05-14 18:04:19

150 degree

150 degree

MathIsFun286
2020-05-14 18:04:19

150 degrees

150 degrees

Patriots_Mathgeeks
2020-05-14 18:04:20

150 degress

150 degress

rrusczyk
2020-05-14 18:04:34

How do we see that $m\angle COD$ is $150$ degrees?

How do we see that $m\angle COD$ is $150$ degrees?

DerpyTaterTot
2020-05-14 18:05:14

its 360-45-45-60-60

its 360-45-45-60-60

yayatheduck
2020-05-14 18:05:14

360-60-60-90

360-60-60-90

MR_67
2020-05-14 18:05:14

360-60-60-45-45

360-60-60-45-45

Jalenluorion
2020-05-14 18:05:14

360-60-60-45-45

360-60-60-45-45

KingRavi
2020-05-14 18:05:14

360-60-60-45-45

360-60-60-45-45

nathanqiu
2020-05-14 18:05:14

360-(60+60+45+45)

360-(60+60+45+45)

MathIsFun286
2020-05-14 18:05:14

360 -60-60-45-45=150

360 -60-60-45-45=150

darksol
2020-05-14 18:05:14

360 - (2*45) - (2*60) = 150

360 - (2*45) - (2*60) = 150

sosiaops
2020-05-14 18:05:14

360-(60+60+45++45)

360-(60+60+45++45)

Awesome3.14
2020-05-14 18:05:14

360-90-60-60=150

360-90-60-60=150

sarasota25
2020-05-14 18:05:14

360-45-45-60-60=150

360-45-45-60-60=150

quagmireradical
2020-05-14 18:05:14

360-90-60-60

360-90-60-60

btc433
2020-05-14 18:05:17

$360-60-60-90=150$

$360-60-60-90=150$

rrusczyk
2020-05-14 18:05:22

We know that $m\angle COA = 60^\circ,$ and similarly $m\angle AOE = 60^\circ.$

We know that $m\angle COA = 60^\circ,$ and similarly $m\angle AOE = 60^\circ.$

rrusczyk
2020-05-14 18:05:23

And we also know that $m\angle DOE = 90^\circ$.

And we also know that $m\angle DOE = 90^\circ$.

rrusczyk
2020-05-14 18:05:24

So we have $m\angle COD = 360^\circ - 90^\circ -60^\circ - 60^\circ = 150^\circ.$

So we have $m\angle COD = 360^\circ - 90^\circ -60^\circ - 60^\circ = 150^\circ.$

rrusczyk
2020-05-14 18:05:42

All right, what's the length of arc $\widehat{CD}$?

All right, what's the length of arc $\widehat{CD}$?

CHIPPER33
2020-05-14 18:06:23

5/12 * circumference

5/12 * circumference

sixoneeight
2020-05-14 18:06:23

so the length is 150/360*20pi=25/3*pi

so the length is 150/360*20pi=25/3*pi

Paulfrank
2020-05-14 18:06:23

100pi/12

100pi/12

tumbleweed
2020-05-14 18:06:23

$\frac{25\pi}{3}$

$\frac{25\pi}{3}$

xMidnightFirex
2020-05-14 18:06:23

100pi/12 = 25pi/3

100pi/12 = 25pi/3

Lux1
2020-05-14 18:06:23

so then the arc would measure 15/36 of the circumference

so then the arc would measure 15/36 of the circumference

Awesome3.14
2020-05-14 18:06:23

150/360 * 20pi

150/360 * 20pi

MR_67
2020-05-14 18:06:23

20pi(150/360)

20pi(150/360)

IAM_TSSXII
2020-05-14 18:06:23

25 pi over 3

25 pi over 3

Xcountry
2020-05-14 18:06:23

we need 5/12 of the circumference

we need 5/12 of the circumference

menlo
2020-05-14 18:06:23

150(100pi)/360

150(100pi)/360

and.peggy
2020-05-14 18:06:23

(150/360)*2*10*pi

(150/360)*2*10*pi

jai123
2020-05-14 18:06:23

150/360 times 20 pi

150/360 times 20 pi

winterrain01
2020-05-14 18:06:23

150/360 * 20pi

150/360 * 20pi

rrusczyk
2020-05-14 18:06:26

We see that the curved portion of the rope is $\dfrac{150^\circ}{360^\circ} = \dfrac{5}{12}$ of the whole circumference of the watermelon.

We see that the curved portion of the rope is $\dfrac{150^\circ}{360^\circ} = \dfrac{5}{12}$ of the whole circumference of the watermelon.

rrusczyk
2020-05-14 18:06:27

Therefore, the curved portion of the rope has length $\dfrac{5}{12}(20\pi)$ cm.

Therefore, the curved portion of the rope has length $\dfrac{5}{12}(20\pi)$ cm.

rrusczyk
2020-05-14 18:06:32

Putting all the pieces of the rope together, its total length is \[10\sqrt{3}+ \frac{5}{12}(20\pi) + 10.\]

Putting all the pieces of the rope together, its total length is \[10\sqrt{3}+ \frac{5}{12}(20\pi) + 10.\]

rrusczyk
2020-05-14 18:06:49

Time to get out your calculator!

Time to get out your calculator!

onedance
2020-05-14 18:06:53

53.5?

53.5?

CHIPPER33
2020-05-14 18:06:53

53.5!!

53.5!!

nathanqiu
2020-05-14 18:06:53

53.5!

53.5!

Awesome3.14
2020-05-14 18:07:04

The length of the rope is 53.5

The length of the rope is 53.5

yukrant1
2020-05-14 18:07:04

53.5!!!

53.5!!!

evanshawn316
2020-05-14 18:07:04

53.5

53.5

bronzetruck2016
2020-05-14 18:07:04

53.5

53.5

tigerjade003
2020-05-14 18:07:04

53.5

53.5

Heidi0902
2020-05-14 18:07:04

53.5

53.5

mohanty
2020-05-14 18:07:04

53.5

53.5

arirah9
2020-05-14 18:07:07

53.5!!!!!

53.5!!!!!

Nefarious
2020-05-14 18:07:07

53.5

53.5

rrusczyk
2020-05-14 18:07:10

Compute this, and you should find that the rope is approximately $\boxed{53.5}$ cm long.

Compute this, and you should find that the rope is approximately $\boxed{53.5}$ cm long.

palindrome868
2020-05-14 18:07:15

yep!!

yep!!

sosiaops
2020-05-14 18:07:15

WE DID IT!!!

WE DID IT!!!

sosiaops
2020-05-14 18:07:24

YAY WE DID IT!

YAY WE DID IT!

advanture
2020-05-14 18:07:24

YAY!

YAY!

rrusczyk
2020-05-14 18:07:39

All right, back to DPatrick!

All right, back to DPatrick!

DPatrick
2020-05-14 18:07:47

Hi again!

Hi again!

DPatrick
2020-05-14 18:07:51

**Target #8:**Tyrell makes a list of the left-most digit of the powers of $2,$ from $2^0$ to $2^{1000},$ inclusive. The first six numbers on Tyrell’s list, therefore, are $1, 2, 4, 8, 1, 3$ and the last six numbers are $3, 6, 1, 2, 5, 1.$ Given that $2^{1000}$ has $302$ digits, and that $8$ appears $52$ times on Tyrell’s list, how many times does $9$ appear in his list?
DPatrick
2020-05-14 18:08:11

Of all 46 problems on all three contests, I think this one was my favorite.

Of all 46 problems on all three contests, I think this one was my favorite.

DPatrick
2020-05-14 18:08:33

We're given a lot of information, but it's all a little hard to use.

We're given a lot of information, but it's all a little hard to use.

DPatrick
2020-05-14 18:08:44

Any ideas?

Any ideas?

Maths4J
2020-05-14 18:09:12

46 is the max score but there are 48 problems, 10+8+30

46 is the max score but there are 48 problems, 10+8+30

aie8920
2020-05-14 18:09:12

Aren't there 48 problems

Aren't there 48 problems

Quaoar
2020-05-14 18:09:17

Weren't there 48 problems?

Weren't there 48 problems?

DPatrick
2020-05-14 18:09:28

Oops, you're right! This one is still my favorite, though.

Oops, you're right! This one is still my favorite, though.

romani
2020-05-14 18:09:53

There is a pattern

There is a pattern

Parentcoach_Maths
2020-05-14 18:09:53

Let’s look for a pattern

Let’s look for a pattern

PShucks
2020-05-14 18:09:53

find a pattern!

find a pattern!

LlamaWarrior
2020-05-14 18:09:53

Find a pattern!

Find a pattern!

PShucks
2020-05-14 18:09:53

find a pattern

find a pattern

Unicorn78
2020-05-14 18:09:53

Look for patterns?

Look for patterns?

Leonard_my_dude
2020-05-14 18:09:53

List some out and hope it repeats i guess

List some out and hope it repeats i guess

mathfun42
2020-05-14 18:09:53

it forms a pattern

it forms a pattern

arirah9
2020-05-14 18:09:53

look for a pattern

look for a pattern

MathyAOP
2020-05-14 18:09:53

Look for a pattern

Look for a pattern

Liakas_42
2020-05-14 18:09:53

find a pattern

find a pattern

Pleaseletmewin
2020-05-14 18:09:53

find a pattern

find a pattern

akpi2
2020-05-14 18:09:53

patterns

patterns

winterrain01
2020-05-14 18:09:53

Just get in and mess around?

Just get in and mess around?

mathbw225
2020-05-14 18:09:53

look for a pattern

look for a pattern

DPatrick
2020-05-14 18:10:01

Yeah, let's look for patterns.

Yeah, let's look for patterns.

DPatrick
2020-05-14 18:10:29

Let's think about how the first digit can change. That is, let's think about what can follow each of the numbers in our list.

Let's think about how the first digit can change. That is, let's think about what can follow each of the numbers in our list.

DPatrick
2020-05-14 18:10:33

For example, what can follow a $1$ in the list?

For example, what can follow a $1$ in the list?

pianoboy
2020-05-14 18:10:56

2,3

2,3

Speedstorm
2020-05-14 18:10:56

2 or 3

2 or 3

mfro24
2020-05-14 18:10:56

2 or 3

2 or 3

awesomeming327.
2020-05-14 18:10:56

2,3

2,3

john0512
2020-05-14 18:10:56

2 or 3

2 or 3

Math5K
2020-05-14 18:10:56

2 or 3

2 or 3

AmpharosX
2020-05-14 18:10:56

2 and 3

2 and 3

geodash2
2020-05-14 18:10:56

2 or 3

2 or 3

Pleaseletmewin
2020-05-14 18:10:56

2 or a 3

2 or a 3

Leonard_my_dude
2020-05-14 18:10:56

2 or 3

2 or 3

DottedCaculator
2020-05-14 18:10:56

2 or 3

2 or 3

MTHJJS
2020-05-14 18:10:56

2 or 3

2 or 3

yukrant1
2020-05-14 18:10:56

a 2 or a 3?

a 2 or a 3?

qsheng
2020-05-14 18:10:56

2 or 3

2 or 3

I-_-I
2020-05-14 18:10:56

2 or 3

2 or 3

DPatrick
2020-05-14 18:11:03

Right. If we double a number that starts with $1$, then the resulting number must start with $2$ or $3.$

Right. If we double a number that starts with $1$, then the resulting number must start with $2$ or $3.$

DPatrick
2020-05-14 18:11:13

You might find thinking about small-length numbers helpful. If we double a number between 100 and 199, the resulting number is between 200 and 398. If we double a number between 1000 and 1999, the resulting number is between 2000 and 3998. And so on, regardless of the number of digits.

You might find thinking about small-length numbers helpful. If we double a number between 100 and 199, the resulting number is between 200 and 398. If we double a number between 1000 and 1999, the resulting number is between 2000 and 3998. And so on, regardless of the number of digits.

DPatrick
2020-05-14 18:11:25

So $1$ must always be followed by $2$ or $3.$

So $1$ must always be followed by $2$ or $3.$

DPatrick
2020-05-14 18:11:37

How about $2$? What number(s) can follow $2$ in our list?

How about $2$? What number(s) can follow $2$ in our list?

millburn2006
2020-05-14 18:11:54

4,5

4,5

fasterthanlight
2020-05-14 18:11:54

4,5

4,5

Math4Life2020
2020-05-14 18:11:54

4 5

4 5

jupiter314
2020-05-14 18:11:54

4 or 5

4 or 5

jbear911
2020-05-14 18:11:54

4 or 5

4 or 5

rjtmagic
2020-05-14 18:11:54

4/5

4/5

punkinpiday
2020-05-14 18:11:54

4 or 5

4 or 5

winner320
2020-05-14 18:11:54

2 must be followed ny 4 or 5

2 must be followed ny 4 or 5

Ninja11
2020-05-14 18:11:54

4 or 5

4 or 5

kdraganov
2020-05-14 18:11:54

4 or 5

4 or 5

superagh
2020-05-14 18:11:54

4, 5

4, 5

mathisfun17
2020-05-14 18:11:54

4,5

4,5

Chesssaga
2020-05-14 18:11:54

4,5

4,5

KSH31415
2020-05-14 18:11:54

4 and 5

4 and 5

casp
2020-05-14 18:11:54

4 or 5

4 or 5

DPatrick
2020-05-14 18:11:58

A $2$ must be followed by a $4$ or a $5$, by the same logic.

A $2$ must be followed by a $4$ or a $5$, by the same logic.

DPatrick
2020-05-14 18:12:13

How about $3$? What can follow $3$?

How about $3$? What can follow $3$?

MathCounts145
2020-05-14 18:12:28

6 or 7

6 or 7

Jerry_Guo
2020-05-14 18:12:28

6 or 7

6 or 7

green_alligator
2020-05-14 18:12:28

6, 7

6, 7

ElNoraa
2020-05-14 18:12:28

6,7

6,7

vcric247
2020-05-14 18:12:28

6/7

6/7

Asterlan
2020-05-14 18:12:28

6 or 7

6 or 7

ohbaby
2020-05-14 18:12:28

6 or 7

6 or 7

TheEpicCarrot7
2020-05-14 18:12:28

6 and 7

6 and 7

felicitas
2020-05-14 18:12:28

6 or 7

6 or 7

ccu700037
2020-05-14 18:12:28

6 or 7

6 or 7

jellybeanchocolate
2020-05-14 18:12:28

6 or 7

6 or 7

acornfirst
2020-05-14 18:12:28

6 or 7

6 or 7

DPatrick
2020-05-14 18:12:33

A $3$ must be followed by a $6$ or a $7.$

A $3$ must be followed by a $6$ or a $7.$

DPatrick
2020-05-14 18:12:38

How about $4$?

How about $4$?

menlo
2020-05-14 18:12:53

8,9

8,9

krishgarg
2020-05-14 18:12:53

8,9

8,9

MathIsFun286
2020-05-14 18:12:53

8,9

8,9

QQMath
2020-05-14 18:12:53

8,9

8,9

gs_2006
2020-05-14 18:12:53

4 is an 8 or a 9

4 is an 8 or a 9

lucaszhou
2020-05-14 18:12:53

8,9

8,9

Gentoo
2020-05-14 18:12:53

8 or 9

8 or 9

pi_is_3.14
2020-05-14 18:12:53

8, 9

8, 9

tumbleweed
2020-05-14 18:12:53

8,9

8,9

SR1234
2020-05-14 18:12:53

8 or 9

8 or 9

vitaRaptor2000
2020-05-14 18:12:53

8 or 9

8 or 9

UnicornForev19
2020-05-14 18:12:53

8 or 9

8 or 9

Floating-Clouds
2020-05-14 18:12:53

8 or 9

8 or 9

DPatrick
2020-05-14 18:12:59

A $4$ must be followed by an $8$ or a $9.$

A $4$ must be followed by an $8$ or a $9.$

DPatrick
2020-05-14 18:13:10

How about $5?$

How about $5?$

Awesome3.14
2020-05-14 18:13:30

only 1

only 1

MR_67
2020-05-14 18:13:30

1

1

cat_maniac_8
2020-05-14 18:13:30

1

1

NCEE
2020-05-14 18:13:30

1

1

TThB0501
2020-05-14 18:13:30

1

1

Zhaom
2020-05-14 18:13:30

1 and only 1

1 and only 1

TMSmathcounts737
2020-05-14 18:13:30

and 5 must be followed with a 1

and 5 must be followed with a 1

Fishareweird
2020-05-14 18:13:30

1

1

NinjaMango
2020-05-14 18:13:30

1

1

FireCurve24
2020-05-14 18:13:30

1

1

DPatrick
2020-05-14 18:13:33

A $5$ is always followed by a $1.$ Why?

A $5$ is always followed by a $1.$ Why?

acornfirst
2020-05-14 18:14:17

because it carries over

because it carries over

Anniboy
2020-05-14 18:14:17

it is the leftmost digit

it is the leftmost digit

Lux1
2020-05-14 18:14:17

because 5*2 is 10

because 5*2 is 10

AmpharosX
2020-05-14 18:14:17

Because new digit!

Because new digit!

IMadeYouReadThis
2020-05-14 18:14:17

$5\cdot2=10$

$5\cdot2=10$

harmonyguan
2020-05-14 18:14:17

cuz 5*2=10

cuz 5*2=10

mmjguitar
2020-05-14 18:14:17

becase it is a new place value

becase it is a new place value

Math5K
2020-05-14 18:14:17

50*2=100 and 59*2=118

50*2=100 and 59*2=118

axzhang
2020-05-14 18:14:17

because it will create a new digit infront

because it will create a new digit infront

Juneybug
2020-05-14 18:14:17

oh yeah, because it turns into either a 10 or 11, both of which start with 1

oh yeah, because it turns into either a 10 or 11, both of which start with 1

Dalar25
2020-05-14 18:14:17

because if you double the number, it will come out to 10 or 11, both of which start with 1

because if you double the number, it will come out to 10 or 11, both of which start with 1

E1CAarush
2020-05-14 18:14:17

11 and 10

11 and 10

DPatrick
2020-05-14 18:14:26

Right: doubling a number with a $5$ will add a digit! For example, doubling a number between $500$ and $599$ gives us a number between $1000$ and $1198$.

Right: doubling a number with a $5$ will add a digit! For example, doubling a number between $500$ and $599$ gives us a number between $1000$ and $1198$.

DPatrick
2020-05-14 18:14:47

So we always go to a number that's got one more digit, and the new number starts with $1$.

So we always go to a number that's got one more digit, and the new number starts with $1$.

DPatrick
2020-05-14 18:14:57

So a $5$ in our list is always followed by a $1$.

So a $5$ in our list is always followed by a $1$.

sixoneeight
2020-05-14 18:15:11

5,6,7,8,9 all give 1!

5,6,7,8,9 all give 1!

Potato2017
2020-05-14 18:15:11

6 is 1, 7 is 1, 8 is 1, 9 is 1.

6 is 1, 7 is 1, 8 is 1, 9 is 1.

Pokemon2
2020-05-14 18:15:11

5-9 always gives a 1

5-9 always gives a 1

Chesssaga
2020-05-14 18:15:11

Same goes for 6-9

Same goes for 6-9

Noam2007
2020-05-14 18:15:11

Same with $6,7,8,$ and $9$ because they all add a new digit.

Same with $6,7,8,$ and $9$ because they all add a new digit.

sixoneeight
2020-05-14 18:15:11

so are 6,7,8, and 9!

so are 6,7,8, and 9!

DPatrick
2020-05-14 18:15:32

Aha! It's the same with $6, 7, 8,$ and $9$: they always are followed by a $1.$

Aha! It's the same with $6, 7, 8,$ and $9$: they always are followed by a $1.$

DPatrick
2020-05-14 18:15:47

So we keep cycling back to $1$, every time we add a digit to the underlying number (the power of $2$).

So we keep cycling back to $1$, every time we add a digit to the underlying number (the power of $2$).

yukrant1
2020-05-14 18:15:54

and going back to 1 restarts the pattern

and going back to 1 restarts the pattern

Noam2007
2020-05-14 18:16:03

That means that $1$'s are the most common.

That means that $1$'s are the most common.

DPatrick
2020-05-14 18:16:10

Aha...$1$'s are the key.

Aha...$1$'s are the key.

DPatrick
2020-05-14 18:16:31

We have a $1$ in our list, and we double for a few steps, and we eventually add a digit to our power of $2$ which gets us back to a new $1$ in our list.

We have a $1$ in our list, and we double for a few steps, and we eventually add a digit to our power of $2$ which gets us back to a new $1$ in our list.

E1CAarush
2020-05-14 18:16:44

We need to find a pattern with ones

We need to find a pattern with ones

DPatrick
2020-05-14 18:17:00

I agree. What are the different possible cycles from one $1$ to the next $1$?

I agree. What are the different possible cycles from one $1$ to the next $1$?

qsheng
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

MR_67
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

Zhaom
2020-05-14 18:17:27

1-2-4-8-1

1-2-4-8-1

Jerry_Guo
2020-05-14 18:17:27

1, 2, 4, 8, 1

1, 2, 4, 8, 1

bobjoebilly
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

casp
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

mmjguitar
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

bacond
2020-05-14 18:17:27

1,2,4,8,1

1,2,4,8,1

DPatrick
2020-05-14 18:17:31

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$. (Indeed, this is how the list starts.)

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$. (Indeed, this is how the list starts.)

DPatrick
2020-05-14 18:17:39

What else?

What else?

Doudou_Chen
2020-05-14 18:17:55

1-2-4-9-1-

1-2-4-9-1-

Lcz
2020-05-14 18:18:03

1 2 4 9 1

1 2 4 9 1

mathfun42
2020-05-14 18:18:03

1, 2, 4, 9, 1

1, 2, 4, 9, 1

DPatrick
2020-05-14 18:18:06

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.

maxben
2020-05-14 18:18:16

1 2 5 1

1 2 5 1

advanture
2020-05-14 18:18:16

1-2-5-1

1-2-5-1

eibc
2020-05-14 18:18:16

1, 2, 5, 1

1, 2, 5, 1

Eng123
2020-05-14 18:18:16

1, 2, 5, 1

1, 2, 5, 1

Speedstorm
2020-05-14 18:18:16

1,2,5,1

1,2,5,1

DPatrick
2020-05-14 18:18:20

We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.

sarahAops2020
2020-05-14 18:18:29

1-3-6-1

1-3-6-1

Jalenluorion
2020-05-14 18:18:29

1361

1361

aidni47
2020-05-14 18:18:29

1, 3, 6, 1

1, 3, 6, 1

CHIPPER33
2020-05-14 18:18:29

1,3,6,1

1,3,6,1

albertzhuan
2020-05-14 18:18:29

1, 3, 6, 1

1, 3, 6, 1

ApraTrip
2020-05-14 18:18:29

1,3,6,1

1,3,6,1

manokid
2020-05-14 18:18:29

1361

1361

DPatrick
2020-05-14 18:18:33

We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.

superagh
2020-05-14 18:18:54

1 3 7 1

1 3 7 1

palindrome868
2020-05-14 18:18:54

1-3-7-1

1-3-7-1

nathanqiu
2020-05-14 18:18:54

1,3,7,1

1,3,7,1

onedance
2020-05-14 18:18:54

1,3,7,1

1,3,7,1

sdattilo2002
2020-05-14 18:18:54

1,3,7,1

1,3,7,1

DPatrick
2020-05-14 18:18:57

We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.

DPatrick
2020-05-14 18:19:21

And that's it: if you look at the rules we discovered for what numbers can follow which, there's are the only cycles from one $1$ to the next.

And that's it: if you look at the rules we discovered for what numbers can follow which, there's are the only cycles from one $1$ to the next.

DPatrick
2020-05-14 18:19:29

Let me list them all out again:

Let me list them all out again:

DPatrick
2020-05-14 18:19:44

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.

We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.

We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.

XTJin
2020-05-14 18:20:04

8 only appears in one of the paths... 9 only appears in one of the paths as well...

8 only appears in one of the paths... 9 only appears in one of the paths as well...

asbodke
2020-05-14 18:20:12

1 8 and 1 9 in total

1 8 and 1 9 in total

Maths4J
2020-05-14 18:20:12

So the sum of all the 8s and 9s equals the number of 4s.

So the sum of all the 8s and 9s equals the number of 4s.

bedwinprusik578
2020-05-14 18:20:23

8 and 9 only APPEAR ONCE

8 and 9 only APPEAR ONCE

DPatrick
2020-05-14 18:20:32

Hmmm...some of the cycles take 4 steps to go between $1$s, and some only take 3 steps.

Hmmm...some of the cycles take 4 steps to go between $1$s, and some only take 3 steps.

DPatrick
2020-05-14 18:20:39

And it's the $4$-step cycles that produce $8$s and $9$s.

And it's the $4$-step cycles that produce $8$s and $9$s.

DPatrick
2020-05-14 18:20:53

Do you see how to put all of these pieces together?

Do you see how to put all of these pieces together?

DPatrick
2020-05-14 18:21:03

There's one more bit of data we haven't used yet. (This is a great problem-solving strategy: if you're stuck, look back and see what you haven't used yet.)

There's one more bit of data we haven't used yet. (This is a great problem-solving strategy: if you're stuck, look back and see what you haven't used yet.)

sixoneeight
2020-05-14 18:21:13

Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 ones.

Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 ones.

superagh
2020-05-14 18:21:13

use the 302 digits information

use the 302 digits information

Maths4J
2020-05-14 18:21:13

We know the number of cycles too, 302

We know the number of cycles too, 302

sixoneeight
2020-05-14 18:21:20

Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 of them

Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 of them

Ninja11
2020-05-14 18:21:26

How does 2^1000 with 302 digits help?

How does 2^1000 with 302 digits help?

Doudou_Chen
2020-05-14 18:21:28

is it the 302 digits part?

is it the 302 digits part?

pi_is_3.14
2020-05-14 18:21:31

$2^100$ has 302 digits

$2^100$ has 302 digits

DPatrick
2020-05-14 18:21:37

We haven't yet used the given fact that $2^{1000}$ has $302$ digits.

We haven't yet used the given fact that $2^{1000}$ has $302$ digits.

DPatrick
2020-05-14 18:21:48

They must have told us that for a reason

They must have told us that for a reason

Maths4J
2020-05-14 18:22:00

actually, it says 2^1000 is a 302 digit number starting with 1, it starts the 302nd cycle

actually, it says 2^1000 is a 302 digit number starting with 1, it starts the 302nd cycle

onedance
2020-05-14 18:22:04

2^1000 has 302 digits?

2^1000 has 302 digits?

fasterthanlight
2020-05-14 18:22:16

so there are 301 cycles

so there are 301 cycles

DPatrick
2020-05-14 18:22:24

Right! From our initial $1$, we have to go through $301$ cycles to get to the final $1$. (We have to add $301$ digits to get from the $1$-digit $2^0$ to the $302$-digit $2^{1000}.)$

Right! From our initial $1$, we have to go through $301$ cycles to get to the final $1$. (We have to add $301$ digits to get from the $1$-digit $2^0$ to the $302$-digit $2^{1000}.)$

DPatrick
2020-05-14 18:22:47

And since we're going from $2^0$ to $2^{1000}$, it'll take us $1000$ steps to do so.

And since we're going from $2^0$ to $2^{1000}$, it'll take us $1000$ steps to do so.

DPatrick
2020-05-14 18:22:58

So if we use $301$ cycles to take $1000$ steps, how many of those cycles are $3$-steps and how many are $4$-steps?

So if we use $301$ cycles to take $1000$ steps, how many of those cycles are $3$-steps and how many are $4$-steps?

DPatrick
2020-05-14 18:24:01

Actually, we don't really care how many are $3$-step, since what we care about (the $9$s) only appear in the $4$-steps.

Actually, we don't really care how many are $3$-step, since what we care about (the $9$s) only appear in the $4$-steps.

palindrome868
2020-05-14 18:24:13

97 4-steps

97 4-steps

advanture
2020-05-14 18:24:13

97 4-steps

97 4-steps

Math5K
2020-05-14 18:24:13

97 4-step

97 4-step

yayy
2020-05-14 18:24:13

97 4 step

97 4 step

Leonard_my_dude
2020-05-14 18:24:13

97 4 steps

97 4 steps

bobjoebilly
2020-05-14 18:24:13

97 4-step

97 4-step

Champion1234
2020-05-14 18:24:13

97 4 steps

97 4 steps

green_alligator
2020-05-14 18:24:13

97 4 steps

97 4 steps

sixoneeight
2020-05-14 18:24:18

x+y=301, 3x+4y=1000

x+y=301, 3x+4y=1000

Mathematician1010
2020-05-14 18:24:18

$3x+4y=1000$

$x+y=301$

$3x+4y=1000$

$x+y=301$

jai123
2020-05-14 18:24:18

x+y=301 and 3x+4y=1000

x+y=301 and 3x+4y=1000

DPatrick
2020-05-14 18:24:36

You could certainly set up a system of equations to solve this.

You could certainly set up a system of equations to solve this.

DPatrick
2020-05-14 18:24:53

Here's how I think to think about it: "If every cycle took 3 steps, we'd only take 903 steps. We need 97 extra steps to make it to 1000."

Here's how I think to think about it: "If every cycle took 3 steps, we'd only take 903 steps. We need 97 extra steps to make it to 1000."

DPatrick
2020-05-14 18:25:05

That is, since $3 \cdot 301 = 903$, we need $1000 - 903 = 97$ "extra" steps to make it to $1000$.

That is, since $3 \cdot 301 = 903$, we need $1000 - 903 = 97$ "extra" steps to make it to $1000$.

DPatrick
2020-05-14 18:25:13

So $97$ of the cycles are $4$-step cycles.

So $97$ of the cycles are $4$-step cycles.

DPatrick
2020-05-14 18:25:15

And now what?

And now what?

advanture
2020-05-14 18:25:35

52 have 8's so 45 have 9's!

52 have 8's so 45 have 9's!

gs_2006
2020-05-14 18:25:35

and now you subtract 52 from 97

and now you subtract 52 from 97

Jomo
2020-05-14 18:25:35

So 97-52=45?

So 97-52=45?

fasterthanlight
2020-05-14 18:25:35

97 4-steps, 52 are 1-2-4-8, so 9 comes out 45 times

97 4-steps, 52 are 1-2-4-8, so 9 comes out 45 times

fasterthanlight
2020-05-14 18:25:35

52 are 1-2-4-8

52 are 1-2-4-8

mfro24
2020-05-14 18:25:35

We subtract 52!

We subtract 52!

E1CAarush
2020-05-14 18:25:35

97-52

97-52

Leonard_my_dude
2020-05-14 18:25:35

97 - 52 = 45

97 - 52 = 45

CHIPPER33
2020-05-14 18:25:35

subract 52

subract 52

kred9
2020-05-14 18:25:35

97-52=45 yay

97-52=45 yay

millburn2006
2020-05-14 18:25:35

97-52=45

97-52=45

The_Better_Samuel
2020-05-14 18:25:35

97-52=45 so 45 times 9 appears

97-52=45 so 45 times 9 appears

PShucks
2020-05-14 18:25:35

subtract 45 out

subtract 45 out

EpicGirl
2020-05-14 18:25:35

45 p's

45 p's

qsheng
2020-05-14 18:25:35

97-52=45

97-52=45

DPatrick
2020-05-14 18:25:42

Aha, we're at the finish line!

Aha, we're at the finish line!

DPatrick
2020-05-14 18:25:46

Each of the $97$ $4$-step cycles contains either an $8$ or a $9$ (but not both).

Each of the $97$ $4$-step cycles contains either an $8$ or a $9$ (but not both).

DPatrick
2020-05-14 18:26:01

But we're told that $52$ of them contain an $8.$ (Yet another bit of data from the problem that we haven't used yet!)

But we're told that $52$ of them contain an $8.$ (Yet another bit of data from the problem that we haven't used yet!)

DPatrick
2020-05-14 18:26:13

Therefore, the other $97 - 52 = \boxed{45}$ of them contain a $9$, and we're done!

Therefore, the other $97 - 52 = \boxed{45}$ of them contain a $9$, and we're done!

DPatrick
2020-05-14 18:26:36

I really liked this problem: all the clues were there, but you had to work pretty hard to put them together to solve the mystery.

I really liked this problem: all the clues were there, but you had to work pretty hard to put them together to solve the mystery.

DPatrick
2020-05-14 18:27:07

We're going to finish up tonight by looking at the final two Team problems.

We're going to finish up tonight by looking at the final two Team problems.

DPatrick
2020-05-14 18:27:18

I know we've been here a while, so if you have to go, that's OK!

I know we've been here a while, so if you have to go, that's OK!

DPatrick
2020-05-14 18:27:29

We'll post a transcript of the entire session on the website when we're done.

We'll post a transcript of the entire session on the website when we're done.

smartguy888
2020-05-14 18:27:44

"rrusczyk:We're not sure! Maybe 90 minutes or so." Reality: 3 hours XD

"rrusczyk:We're not sure! Maybe 90 minutes or so." Reality: 3 hours XD

DPatrick
2020-05-14 18:27:59

Yeah, we were a little off. We've never done a Math Jam like this before!

Yeah, we were a little off. We've never done a Math Jam like this before!

rrusczyk
2020-05-14 18:28:02

Harvey has been mocking me for an hour and a half.

Harvey has been mocking me for an hour and a half.

rrusczyk
2020-05-14 18:28:10

Please don't give him any more ammunition.

Please don't give him any more ammunition.

rrusczyk
2020-05-14 18:28:22

Meanwhile, the Other Guy told me he would have finished 2 hours ago.

Meanwhile, the Other Guy told me he would have finished 2 hours ago.

DPatrick
2020-05-14 18:28:33

This has blown away our record for number of people at a Math Jam. (I believe the old record was a little over 800. We had over 1300 today.)

This has blown away our record for number of people at a Math Jam. (I believe the old record was a little over 800. We had over 1300 today.)

DPatrick
2020-05-14 18:28:53

So let's look at Team #9 next:

So let's look at Team #9 next:

DPatrick
2020-05-14 18:28:58

**Team #9:**A cheerleading squad has $14$ cheerleaders, each a different height. How many ways are there for the cheerleaders to line up for a photo in two rows with seven people each, so that each cheerleader in the back row is taller than the one immediately in front of them, and so that the heights of the cheerleaders in the back row descend from the middle to each side?
Quaoar
2020-05-14 18:29:27

Ummmm...

Ummmm...

CT17
2020-05-14 18:29:27

Wow

Wow

DPatrick
2020-05-14 18:29:42

Those were my first thoughts too.

Those were my first thoughts too.

SharonW
2020-05-14 18:29:53

thats a lot of cheerleaders

thats a lot of cheerleaders

sdattilo2002
2020-05-14 18:29:53

that's complicated

that's complicated

mewto
2020-05-14 18:29:53

that looks complicated

that looks complicated

DPatrick
2020-05-14 18:30:05

We have a lot of information to try to organize.

We have a lot of information to try to organize.

winterrain01
2020-05-14 18:30:18

Just assign the cheerleaders heights of 1, 2, 3, ..., 14!

Just assign the cheerleaders heights of 1, 2, 3, ..., 14!

DPatrick
2020-05-14 18:30:35

Great idea! So that they're easier to talk about, let's number the cheerleaders from $1$ to $14$ in order of height, with $1$ being the shortest and $14$ being the tallest.

Great idea! So that they're easier to talk about, let's number the cheerleaders from $1$ to $14$ in order of height, with $1$ being the shortest and $14$ being the tallest.

mahaler
2020-05-14 18:31:10

I think first we should draw it out

I think first we should draw it out

TMSmathcounts737
2020-05-14 18:31:10

Let's start to draw a picture

Let's start to draw a picture

DPatrick
2020-05-14 18:31:27

I wholeheartedly agree.

I wholeheartedly agree.

DPatrick
2020-05-14 18:31:32

A great problem solving strategy is to draw a picture.

A great problem solving strategy is to draw a picture.

DPatrick
2020-05-14 18:31:36

We can convert the text of the problem statement into an easier-to-use diagram:

We can convert the text of the problem statement into an easier-to-use diagram:

DPatrick
2020-05-14 18:31:39

DPatrick
2020-05-14 18:31:53

The $<$ signs are the usual greater-than signs: they indicate where a cheerleader must have a larger number (that is, be taller) than the cheerleader standing next to him or her.

The $<$ signs are the usual greater-than signs: they indicate where a cheerleader must have a larger number (that is, be taller) than the cheerleader standing next to him or her.

DPatrick
2020-05-14 18:32:02

(Or less-than signs, I guess!)

(Or less-than signs, I guess!)

DPatrick
2020-05-14 18:32:10

So we need to count the number of ways we can fill the boxes with the numbers $1$ to $14$, preserving all the inequalities shown.

So we need to count the number of ways we can fill the boxes with the numbers $1$ to $14$, preserving all the inequalities shown.

DPatrick
2020-05-14 18:32:21

Notice anything quick?

Notice anything quick?

Rsar12
2020-05-14 18:32:44

14 must be in the back middle

14 must be in the back middle

Streaks123
2020-05-14 18:32:44

14 must go in center back.

14 must go in center back.

PrinceRoyale
2020-05-14 18:32:44

tallest(14) has to take the back middle spot, because he is the tallest

tallest(14) has to take the back middle spot, because he is the tallest

MR_67
2020-05-14 18:32:44

The middle back person is the tallest

The middle back person is the tallest

albertedwin
2020-05-14 18:32:44

c#14 has to be in the middle of the back row.

c#14 has to be in the middle of the back row.

fasterthanlight
2020-05-14 18:32:44

14 is in the middle position of the back row.

14 is in the middle position of the back row.

maxben
2020-05-14 18:32:44

14 should be in te middle of the second row.

14 should be in te middle of the second row.

IMadeYouReadThis
2020-05-14 18:32:44

$14$ has to be in the back center

$14$ has to be in the back center

nathanqiu
2020-05-14 18:32:44

so middle back is 14

so middle back is 14

EulerRocks2.718
2020-05-14 18:32:44

14 must go in the top middle

14 must go in the top middle

advanture
2020-05-14 18:32:44

14 HAS to be in the middle of the back row

14 HAS to be in the middle of the back row

bobjoebilly
2020-05-14 18:32:44

The one in the middle of the back row needs to be #14.

The one in the middle of the back row needs to be #14.

austinchen2005
2020-05-14 18:32:44

14 is in the middle!

14 is in the middle!

DPatrick
2020-05-14 18:32:55

Yeah! We notice that the $14$ must be in the center of the back row, as this is the only box in the diagram that is not less than some other box.

Yeah! We notice that the $14$ must be in the center of the back row, as this is the only box in the diagram that is not less than some other box.

DPatrick
2020-05-14 18:33:03

DPatrick
2020-05-14 18:33:15

Good start! Always good to get the easiest stuff out of the way.

Good start! Always good to get the easiest stuff out of the way.

DPatrick
2020-05-14 18:33:17

Now what?

Now what?

jupiter314
2020-05-14 18:34:07

there are now 3 ways for 13. either side of 14 or in front of 14

there are now 3 ways for 13. either side of 14 or in front of 14

edjar
2020-05-14 18:34:07

3 choices for 13

3 choices for 13

MathJams
2020-05-14 18:34:07

13 could go in 3 places

13 could go in 3 places

cat_maniac_8
2020-05-14 18:34:07

13 has to be surrounding it

13 has to be surrounding it

bacond
2020-05-14 18:34:07

The 13 has to be on the side of the 14

The 13 has to be on the side of the 14

kkomma10
2020-05-14 18:34:07

the next one is 13

the next one is 13

yayy
2020-05-14 18:34:07

13 is next to it

13 is next to it

MAthen07
2020-05-14 18:34:07

13 has to be next to 14.

13 has to be next to 14.

DPatrick
2020-05-14 18:34:34

I agree: given what's left, the $13$ has to either be in the back row next to $14$, or could be directly in front of $14$.

I agree: given what's left, the $13$ has to either be in the back row next to $14$, or could be directly in front of $14$.

DPatrick
2020-05-14 18:34:52

Every other box is shorter than another box that's still empty.

Every other box is shorter than another box that's still empty.

DPatrick
2020-05-14 18:35:10

This feels like that this sort of "constructive counting" is starting to get ugly.

This feels like that this sort of "constructive counting" is starting to get ugly.

DPatrick
2020-05-14 18:35:36

In counting problems, I often like to step back and take a look at the big picture before diving into too much casework.

In counting problems, I often like to step back and take a look at the big picture before diving into too much casework.

DPatrick
2020-05-14 18:35:54

Do you have any "big picture" observations about our diagram?

Do you have any "big picture" observations about our diagram?

DPatrick
2020-05-14 18:35:58

harmonyguan
2020-05-14 18:36:18

symmetrical?

symmetrical?

pi_is_3.14
2020-05-14 18:36:23

symmetry

symmetry

Eng123
2020-05-14 18:36:23

Symmetry.

Symmetry.

DPatrick
2020-05-14 18:36:41

Indeed: the left side and the right side look like mirror images of the same picture.

Indeed: the left side and the right side look like mirror images of the same picture.

PrinceRoyale
2020-05-14 18:37:03

you can cut it in half and solve one side only. but remember to multiply at the end

you can cut it in half and solve one side only. but remember to multiply at the end

DPatrick
2020-05-14 18:37:23

Aha -- it looks like the left side and the right side are essentially the same.

Aha -- it looks like the left side and the right side are essentially the same.

sixoneeight
2020-05-14 18:37:28

First, you have to choose the one in front of 14

First, you have to choose the one in front of 14

DPatrick
2020-05-14 18:37:40

True, let's not forget about that too!

True, let's not forget about that too!

DPatrick
2020-05-14 18:37:57

So it looks like we've observed that the diagram splits into three pieces from here:

So it looks like we've observed that the diagram splits into three pieces from here:

DPatrick
2020-05-14 18:38:01

DPatrick
2020-05-14 18:38:17

That is, the remaining cheerleaders can be divided into three groups: $6$ to go in the three columns on the left side shaded in blue, $6$ to go in the three columns on the right side shaded in green, and the thirteenth to go in the center of the front row shaded in pink.

That is, the remaining cheerleaders can be divided into three groups: $6$ to go in the three columns on the left side shaded in blue, $6$ to go in the three columns on the right side shaded in green, and the thirteenth to go in the center of the front row shaded in pink.

casp
2020-05-14 18:38:40

there are 13 ways

there are 13 ways

yangi26
2020-05-14 18:38:40

you have 13 choices for the one in front of 14

you have 13 choices for the one in front of 14

nathanqiu
2020-05-14 18:38:40

13 choices for pink

13 choices for pink

jupiter314
2020-05-14 18:38:40

there are 13 ways to choose the person in front.

there are 13 ways to choose the person in front.

DPatrick
2020-05-14 18:39:00

Good, let's do that next: we have $13$ choices for who goes in the pink spot in front of $14$. Anybody can go there!

Good, let's do that next: we have $13$ choices for who goes in the pink spot in front of $14$. Anybody can go there!

DPatrick
2020-05-14 18:39:10

So now we have twelve of them left.

So now we have twelve of them left.

DPatrick
2020-05-14 18:39:21

How do we split them into two groups of six, to go on either side?

How do we split them into two groups of six, to go on either side?

sixoneeight
2020-05-14 18:40:00

12 choose 6

12 choose 6

krishgarg
2020-05-14 18:40:00

12 choose 6

12 choose 6

kred9
2020-05-14 18:40:00

12C6?

12C6?

kdraganov
2020-05-14 18:40:00

$\dbinom{12}{6}$

$\dbinom{12}{6}$

Eng123
2020-05-14 18:40:00

12 choose 6.

12 choose 6.

darthtator012
2020-05-14 18:40:00

12 choose 6

12 choose 6

Rsar12
2020-05-14 18:40:00

12 choose 6?

12 choose 6?

mohanty
2020-05-14 18:40:00

we pick any 6 out of 12

we pick any 6 out of 12

mfro24
2020-05-14 18:40:00

We have 12c6 ways to do it!

We have 12c6 ways to do it!

Alculator11
2020-05-14 18:40:00

$\binom{12}{6}$

$\binom{12}{6}$

MR_67
2020-05-14 18:40:00

12C6

12C6

DPatrick
2020-05-14 18:40:17

Right: this is a job for combinations!

Right: this is a job for combinations!

DPatrick
2020-05-14 18:40:45

We can choose $6$ of the $12$ to go on the blue side in $\dbinom{12}{6}$ ways.

We can choose $6$ of the $12$ to go on the blue side in $\dbinom{12}{6}$ ways.

DPatrick
2020-05-14 18:40:57

Then the remaining $6$ will go on the green side.

Then the remaining $6$ will go on the green side.

fasterthanlight
2020-05-14 18:41:11

924 ways

924 ways

DarrenHuang888
2020-05-14 18:41:11

924 is 12 choose 6

924 is 12 choose 6

mmjguitar
2020-05-14 18:41:11

924

924

Beastboss
2020-05-14 18:41:11

924 ways

924 ways

manokid
2020-05-14 18:41:11

924

924

DPatrick
2020-05-14 18:41:22

Right: good thing that calculator is still handy!

Right: good thing that calculator is still handy!

DPatrick
2020-05-14 18:41:28

$\dbinom{12}{6} = 924$

$\dbinom{12}{6} = 924$

DPatrick
2020-05-14 18:41:51

So to recap the choices so far: we had $13$ choices for the pink, then $924$ choices for the group of $6$ to be blue, and the rest are green.

So to recap the choices so far: we had $13$ choices for the pink, then $924$ choices for the group of $6$ to be blue, and the rest are green.

advanture
2020-05-14 18:42:03

then you order them!

then you order them!

DPatrick
2020-05-14 18:42:23

Right, there's still more do to! For each of the groups of $6$ to one side, we have to count the number of ways they can be arranged.

Right, there's still more do to! For each of the groups of $6$ to one side, we have to count the number of ways they can be arranged.

DPatrick
2020-05-14 18:42:49

Since only the order of their heights matter, let's relabel them $A$ (shortest) through $F$ (tallest), just so it's easier to keep track of them.

Since only the order of their heights matter, let's relabel them $A$ (shortest) through $F$ (tallest), just so it's easier to keep track of them.

DPatrick
2020-05-14 18:43:05

Oops

Oops

DPatrick
2020-05-14 18:43:08

So for the left side, we need to count the ways to place $A$ through $F$ in the diagram below, such that all of the inequality signs work:

So for the left side, we need to count the ways to place $A$ through $F$ in the diagram below, such that all of the inequality signs work:

DPatrick
2020-05-14 18:43:12

cj13609517288
2020-05-14 18:43:36

Hey, we know where to put the greatest number in the 6-group!

Hey, we know where to put the greatest number in the 6-group!

dolphin7
2020-05-14 18:43:36

F must go next to 14

F must go next to 14

Alculator11
2020-05-14 18:43:36

F goes in the top right

F goes in the top right

Awesome3.14
2020-05-14 18:43:36

F goes on the top right

F goes on the top right

Quaoar
2020-05-14 18:43:36

Tallest in the top right

Tallest in the top right

cj13609517288
2020-05-14 18:43:36

F goes in top right

F goes in top right

vtlev
2020-05-14 18:43:36

F is in op right

F is in op right

E1CAarush
2020-05-14 18:43:36

The top right is the greatest

The top right is the greatest

bobjoebilly
2020-05-14 18:43:36

F must be in the upper right hand corner.

F must be in the upper right hand corner.

FearlessTaurus
2020-05-14 18:43:36

F goes in the top right

F goes in the top right

Speedstorm
2020-05-14 18:43:36

F is in the top right

F is in the top right

DPatrick
2020-05-14 18:43:46

Good: we see that $F$ must go in the upper-right box, as every other box is less than some other box.

Good: we see that $F$ must go in the upper-right box, as every other box is less than some other box.

DPatrick
2020-05-14 18:43:49

cat_maniac_8
2020-05-14 18:44:21

E is either bellow are to the left of F

E is either bellow are to the left of F

Kruxe
2020-05-14 18:44:21

then E could go in the one to the left or the one underneath

then E could go in the one to the left or the one underneath

melonlord
2020-05-14 18:44:21

E is in the top middle or bottom right

E is in the top middle or bottom right

sri1priya
2020-05-14 18:44:21

E has 2 choices?

E has 2 choices?

sigma_notation
2020-05-14 18:44:21

E must go next to F

E must go next to F

peace09
2020-05-14 18:44:21

E is next to F or below F

E is next to F or below F

DPatrick
2020-05-14 18:44:29

We see that $E$ must either go in the lower-right box or the upper-center box.

We see that $E$ must either go in the lower-right box or the upper-center box.

DPatrick
2020-05-14 18:44:44

I don't think there's anything much clever we can do except to count the two cases.

I don't think there's anything much clever we can do except to count the two cases.

DPatrick
2020-05-14 18:44:54

Fortunately the diagram is small, so hopefully it'll not be too bad.

Fortunately the diagram is small, so hopefully it'll not be too bad.

DPatrick
2020-05-14 18:45:00

Case 1: we place $E$ in the lower-right box:

Case 1: we place $E$ in the lower-right box:

DPatrick
2020-05-14 18:45:03

DPatrick
2020-05-14 18:45:14

How do we finish with A-D?

How do we finish with A-D?

Alculator11
2020-05-14 18:45:40

Now we have 1 choice for D

Now we have 1 choice for D

nathanqiu
2020-05-14 18:45:40

D is left of F

D is left of F

Gentoo
2020-05-14 18:45:40

Then D has to go next to F

Then D has to go next to F

sdattilo2002
2020-05-14 18:45:40

d goes next to f

d goes next to f

tigerjade003
2020-05-14 18:45:40

D must be right next to F

D must be right next to F

PShucks
2020-05-14 18:45:40

d must be top center

d must be top center

DPatrick
2020-05-14 18:45:45

Then $D$ must go in the upper-center box.

Then $D$ must go in the upper-center box.

DPatrick
2020-05-14 18:45:48

DPatrick
2020-05-14 18:46:11

How do we finish this case?

How do we finish this case?

melonlord
2020-05-14 18:46:23

C is in the top left or bottom middle

C is in the top left or bottom middle

jupiter314
2020-05-14 18:46:23

There are 2 choices for C

There are 2 choices for C

cat_maniac_8
2020-05-14 18:46:23

C goes either below or to the left of D

C goes either below or to the left of D

MR_67
2020-05-14 18:46:23

C goes either middle bottom or top left

C goes either middle bottom or top left

Alculator11
2020-05-14 18:46:30

Now 2 choices for C

Now 2 choices for C

manokid
2020-05-14 18:46:30

Then c can go below or next to d

Then c can go below or next to d

Unicorn78
2020-05-14 18:46:30

C has two choices

C has two choices

BakedPotato66
2020-05-14 18:46:30

C can either go to the left or in front of D

C can either go to the left or in front of D

DPatrick
2020-05-14 18:46:35

Right.

Right.

DPatrick
2020-05-14 18:46:53

We could put $C$ in the bottom-middle, and then $A$ and $B$ are forced:

We could put $C$ in the bottom-middle, and then $A$ and $B$ are forced:

DPatrick
2020-05-14 18:46:57

DPatrick
2020-05-14 18:47:08

Or else $C$ goes in the upper-left, in which case $A$ and $B$ can fill the remaining two spots in either order:

Or else $C$ goes in the upper-left, in which case $A$ and $B$ can fill the remaining two spots in either order:

DPatrick
2020-05-14 18:47:12

DPatrick
2020-05-14 18:47:16

DPatrick
2020-05-14 18:47:38

So that's $3$ ways to place the group of six in this case.

So that's $3$ ways to place the group of six in this case.

DarrenHuang888
2020-05-14 18:47:46

we could have E next to F too

we could have E next to F too

DPatrick
2020-05-14 18:47:55

Right, the other case was that $E$ ends up next to $F$:

Right, the other case was that $E$ ends up next to $F$:

DPatrick
2020-05-14 18:47:59

DPatrick
2020-05-14 18:48:11

How do we finish counting this case from here?

How do we finish counting this case from here?

DPatrick
2020-05-14 18:48:48

There are lots of cases for $D$, but is there a simpler way to count this case?

There are lots of cases for $D$, but is there a simpler way to count this case?

cj13609517288
2020-05-14 18:49:02

24(total)/2(only half work for the left)=12

24(total)/2(only half work for the left)=12

DPatrick
2020-05-14 18:49:17

That's a great observation!

That's a great observation!

DPatrick
2020-05-14 18:49:33

There are $4! = 24$ ways to place $A$ through $D$ in any fashion.

There are $4! = 24$ ways to place $A$ through $D$ in any fashion.

DPatrick
2020-05-14 18:49:56

But exactly half of them work in the first column, with a taller person in back. By symmetry, the other half illegally have a taller person in front.

But exactly half of them work in the first column, with a taller person in back. By symmetry, the other half illegally have a taller person in front.

DPatrick
2020-05-14 18:50:03

So there are $4!/2 = 12$ ways to finish this case.

So there are $4!/2 = 12$ ways to finish this case.

DPatrick
2020-05-14 18:50:16

Adding the cases for the group of $6$ on the left, how many ways to arrange them?

Adding the cases for the group of $6$ on the left, how many ways to arrange them?

TheEpicCarrot7
2020-05-14 18:50:33

3+12=15

3+12=15

Jomo
2020-05-14 18:50:33

15

15

edjar
2020-05-14 18:50:33

15

15

dolphin7
2020-05-14 18:50:33

15

15

winner320
2020-05-14 18:50:33

15

15

Zhaom
2020-05-14 18:50:33

15

15

DPatrick
2020-05-14 18:50:37

Adding our two cases gives us $3+12 = 15$ ways to arrange the cheerleaders on the left side.

Adding our two cases gives us $3+12 = 15$ ways to arrange the cheerleaders on the left side.

Quaoar
2020-05-14 18:51:02

same for right side

same for right side

edjar
2020-05-14 18:51:02

And 15 more on the right

And 15 more on the right

karthic7073
2020-05-14 18:51:02

And it would be the same for the other side!

And it would be the same for the other side!

Speedstorm
2020-05-14 18:51:02

so there's 15 ways to arrange them on the right side as well

so there's 15 ways to arrange them on the right side as well

DPatrick
2020-05-14 18:51:12

Right, don't forget! Since the right side is exactly the same (it's just a mirror image of the left side), there are also $15$ ways to arrange the cheerleaders on the left side.

Right, don't forget! Since the right side is exactly the same (it's just a mirror image of the left side), there are also $15$ ways to arrange the cheerleaders on the left side.

mfro24
2020-05-14 18:51:23

So we have a total of $13\cdot\binom{12}{6}\cdot(15^2)$, right?

So we have a total of $13\cdot\binom{12}{6}\cdot(15^2)$, right?

DPatrick
2020-05-14 18:51:44

Exactly. If we go back and look at all the choices we had along the way:

Exactly. If we go back and look at all the choices we had along the way:

DPatrick
2020-05-14 18:51:51

We had $13$ choices for the pink person,

We had $13$ choices for the pink person,

DPatrick
2020-05-14 18:52:02

We had $\dbinom{12}{6}$ choices for the split into two groups of six.

We had $\dbinom{12}{6}$ choices for the split into two groups of six.

DPatrick
2020-05-14 18:52:09

And we had $15$ choices for each of the two groups.

And we had $15$ choices for each of the two groups.

DPatrick
2020-05-14 18:52:23

So the total count is $13 \cdot \dbinom{12}{6} \cdot 15^2$

So the total count is $13 \cdot \dbinom{12}{6} \cdot 15^2$

fasterthanlight
2020-05-14 18:52:40

2,702,700 ways!

2,702,700 ways!

readandartabc
2020-05-14 18:52:40

so there are 2702700 number of cases total

so there are 2702700 number of cases total

fasterthanlight
2020-05-14 18:52:40

2,702,700 ways to arrange the cheerleaders

2,702,700 ways to arrange the cheerleaders

mfro24
2020-05-14 18:52:40

Which gives us a grand total of $2,702,700$!

Which gives us a grand total of $2,702,700$!

FSMS-BL
2020-05-14 18:52:40

now we just put it into a calculator

now we just put it into a calculator

MR_67
2020-05-14 18:52:40

Answer: 13*12C6*15*!5 = 2702700

Answer: 13*12C6*15*!5 = 2702700

rneelam
2020-05-14 18:52:40

$2702700$ is the answer!

$2702700$ is the answer!

MR_67
2020-05-14 18:52:40

2702700

2702700

Jomo
2020-05-14 18:52:40

$$13 \times 924 \times 15^2 = 2,702,700$$

$$13 \times 924 \times 15^2 = 2,702,700$$

manokid
2020-05-14 18:52:40

2702700

2702700

DPatrick
2020-05-14 18:52:46

Therefore, our final count is $12012 \cdot 225 = \boxed{2702700}$ (ways).

Therefore, our final count is $12012 \cdot 225 = \boxed{2702700}$ (ways).

winterrain01
2020-05-14 18:52:58

A nice number

A nice number

Awesome3.14
2020-05-14 18:52:58

wow that is a lot!

wow that is a lot!

DPatrick
2020-05-14 18:53:22

OK, back to rrusczyk for Team #10!

OK, back to rrusczyk for Team #10!

sdattilo2002
2020-05-14 18:53:39

last one!

last one!

MathAcademyIsFun
2020-05-14 18:53:39

is this the last one

is this the last one

rrusczyk
2020-05-14 18:53:44

Last problem for today!

Last problem for today!

rrusczyk
2020-05-14 18:53:48

Thanks for staying with us!

Thanks for staying with us!

Alculator11
2020-05-14 18:53:53

@rrusczyk Why isn't Harvey in this Math Jam?

@rrusczyk Why isn't Harvey in this Math Jam?

rrusczyk
2020-05-14 18:53:59

But he is! Don't you see him?

But he is! Don't you see him?

Jerry_Guo
2020-05-14 18:54:41

does harvey exist?

does harvey exist?

rrusczyk
2020-05-14 18:54:45

He sure thinks he does.

He sure thinks he does.

ohjoshuaoh
2020-05-14 18:54:48

harvey is the best, no offense Mr. Rusczyk

harvey is the best, no offense Mr. Rusczyk

rrusczyk
2020-05-14 18:54:57

None taken. Harvey is pretty awesome.

None taken. Harvey is pretty awesome.

rrusczyk
2020-05-14 18:55:01

Now, time for the last problem.

Now, time for the last problem.

rrusczyk
2020-05-14 18:55:10

Thanks for staying with us for so long. Nearly 4 hours.

Thanks for staying with us for so long. Nearly 4 hours.

rrusczyk
2020-05-14 18:55:15

And still over 700 people here!

And still over 700 people here!

rrusczyk
2020-05-14 18:55:27

**Team #10:**If six numbers are chosen at random, with replacement, from the set of integers from $1$ to $900,$ inclusive, what is the probability that the product of these six integers leaves a remainder of $4$ when divided by $30?$ Express your answer as a percent to the nearest thousandth.
rrusczyk
2020-05-14 18:55:32

Let's start off by calling our product $P$ so we can talk about it.

Let's start off by calling our product $P$ so we can talk about it.

rrusczyk
2020-05-14 18:55:46

What does "$P$ leaves a remainder of $4$ when divided by $30$" tell us about $P$ that might be helpful?

What does "$P$ leaves a remainder of $4$ when divided by $30$" tell us about $P$ that might be helpful?

Jomo
2020-05-14 18:56:01

It must be divisible by 2.

It must be divisible by 2.

cj13609517288
2020-05-14 18:56:01

P is even.

P is even.

Speedstorm
2020-05-14 18:56:01

It's even

It's even

Derpy123
2020-05-14 18:56:01

p is even

p is even

Eng123
2020-05-14 18:56:02

Even.

Even.

rrusczyk
2020-05-14 18:56:07

Well, for starters, $P$ is clearly even. "$P$ leaves a remainder of $4$ when divided by $30$" means we can write $P$ as $30k+4$ for some integer $k.$

Well, for starters, $P$ is clearly even. "$P$ leaves a remainder of $4$ when divided by $30$" means we can write $P$ as $30k+4$ for some integer $k.$

rrusczyk
2020-05-14 18:56:08

So, $P = 30k+4 = 2(15k+2),$ which means $P$ is even.

So, $P = 30k+4 = 2(15k+2),$ which means $P$ is even.

rocketsri
2020-05-14 18:56:19

P is 4 mod 30

P is 4 mod 30

Alculator11
2020-05-14 18:56:19

$P\equiv 4\mod 30$

$P\equiv 4\mod 30$

Notaval
2020-05-14 18:56:19

4 mod 30

4 mod 30

rrusczyk
2020-05-14 18:56:57

A lot of you are talking about mods. That's great! Let's see how far we can get without ever using modular arithmetic -- a lot of MATHCOUNTS students haven't learned modular arithmetic yet. (If you haven't, go check it out.)

A lot of you are talking about mods. That's great! Let's see how far we can get without ever using modular arithmetic -- a lot of MATHCOUNTS students haven't learned modular arithmetic yet. (If you haven't, go check it out.)

rrusczyk
2020-05-14 18:57:04

So, $P$ is even.

So, $P$ is even.

rrusczyk
2020-05-14 18:57:06

What else?

What else?

thanosaops
2020-05-14 18:57:25

P is not divisible by either 3 or 5

P is not divisible by either 3 or 5

Notaval
2020-05-14 18:57:25

not divisible by 3

not divisible by 3

MR_67
2020-05-14 18:57:25

P is not both a multiple of 3 and 5

P is not both a multiple of 3 and 5

thanosaops
2020-05-14 18:57:25

P is not divisible by 3 or 5

P is not divisible by 3 or 5

Jomo
2020-05-14 18:57:25

P is not divisible by 3.

P is not divisible by 3.

MathJams
2020-05-14 18:57:40

P is not a multiple of 3 or 5

P is not a multiple of 3 or 5

rrusczyk
2020-05-14 18:57:54

It's not a multiple of 3 or 5. Could it be any odd that's not a multiple of 3 or 5?

It's not a multiple of 3 or 5. Could it be any odd that's not a multiple of 3 or 5?

kylewu0715
2020-05-14 18:58:11

seven

seven

tigerjade003
2020-05-14 18:58:20

NO IT HAS TO BE EVEN!

NO IT HAS TO BE EVEN!

rrusczyk
2020-05-14 18:58:30

Oops, I mean could it be any even that's not a multiple of 3 or 5?

Oops, I mean could it be any even that's not a multiple of 3 or 5?

BakedPotato66
2020-05-14 18:58:51

8

8

cj13609517288
2020-05-14 18:58:54

it can't, like 8 surpasses

it can't, like 8 surpasses

sigma_notation
2020-05-14 18:58:57

8 doesn't work, so no

8 doesn't work, so no

rrusczyk
2020-05-14 18:59:01

Can't be 8 either.

Can't be 8 either.

rrusczyk
2020-05-14 18:59:11

Why not?

Why not?

Lux1
2020-05-14 18:59:18

it has to leave a remainder of 4 when divided by 30

it has to leave a remainder of 4 when divided by 30

Dalar25
2020-05-14 18:59:27

it has to have a remainder of 4

it has to have a remainder of 4

winner320
2020-05-14 18:59:27

it has to have remainder 4

it has to have remainder 4

rrusczyk
2020-05-14 18:59:28

Great.

Great.

rrusczyk
2020-05-14 18:59:40

So, we've already seen what must happen if we divide by 2.

So, we've already seen what must happen if we divide by 2.

rrusczyk
2020-05-14 18:59:53

We've seen that we can't get an integer if we divide by 3.

We've seen that we can't get an integer if we divide by 3.

rrusczyk
2020-05-14 19:00:05

Can we figure anything else out about dividing by 3?

Can we figure anything else out about dividing by 3?

yayatheduck
2020-05-14 19:00:28

remainder 1

remainder 1

MathJams
2020-05-14 19:00:28

leaves a remainder of 1

leaves a remainder of 1

sixoneeight
2020-05-14 19:00:28

remainder 1

remainder 1

Rsar12
2020-05-14 19:00:28

leaves a remainder of 1

leaves a remainder of 1

MAthen07
2020-05-14 19:00:28

Remainder 1

Remainder 1

Alculator11
2020-05-14 19:00:28

We get a remainder of 1

We get a remainder of 1

mfro24
2020-05-14 19:00:28

It has a remainder of 1!

It has a remainder of 1!

maxben
2020-05-14 19:00:28

remainder is 1

remainder is 1

edjar
2020-05-14 19:00:28

must be 1 more than multiple of 3

must be 1 more than multiple of 3

rrusczyk
2020-05-14 19:00:45

Exactly. For those of you new to "modular arithmetic", we say:

Exactly. For those of you new to "modular arithmetic", we say:

Alculator11
2020-05-14 19:00:48

It's 1 mod 3

It's 1 mod 3

rrusczyk
2020-05-14 19:01:00

That's all "1 mod 3" means -- 1 more than a multiple or 3.

That's all "1 mod 3" means -- 1 more than a multiple or 3.

rrusczyk
2020-05-14 19:01:04

$P$ must be one more than a multiple of $3,$ since \[P=30k+4=30k+3+1=3(10k+1)+1.\]

$P$ must be one more than a multiple of $3,$ since \[P=30k+4=30k+3+1=3(10k+1)+1.\]

rrusczyk
2020-05-14 19:01:13

All right, and what happens if we divide by 5?

All right, and what happens if we divide by 5?

MathJams
2020-05-14 19:01:37

remainder of 4

remainder of 4

advanture
2020-05-14 19:01:37

remain 4

remain 4

rocketsri
2020-05-14 19:01:37

Remainder 4

Remainder 4

Math5K
2020-05-14 19:01:37

remainder 4

remainder 4

Jomo
2020-05-14 19:01:37

4 more.

4 more.

MAthen07
2020-05-14 19:01:37

Remainder 4

Remainder 4

TheEpicCarrot7
2020-05-14 19:01:37

4 as a remainder

4 as a remainder

advanture
2020-05-14 19:01:37

remainder 4

remainder 4

yayatheduck
2020-05-14 19:01:37

remainder 4

remainder 4

DarrenHuang888
2020-05-14 19:01:37

remainder of 4

remainder of 4

manokid
2020-05-14 19:01:37

Remainder of 4

Remainder of 4

millburn2006
2020-05-14 19:01:37

remainder of 4

remainder of 4

rrusczyk
2020-05-14 19:01:40

$P$ must be $4$ more than a multiple of $5,$ since $P = 30k+4 = 5(6k)+4.$

$P$ must be $4$ more than a multiple of $5,$ since $P = 30k+4 = 5(6k)+4.$

rrusczyk
2020-05-14 19:01:49

Another way to say this:

Another way to say this:

sdattilo2002
2020-05-14 19:01:52

it's 4 mod 5

it's 4 mod 5

jupiter314
2020-05-14 19:01:52

it's 4 mod 5

it's 4 mod 5

rrusczyk
2020-05-14 19:01:55

Now...

Now...

rrusczyk
2020-05-14 19:01:57

Why are we focusing on $2, 3,$ and $5$ here?

Why are we focusing on $2, 3,$ and $5$ here?

Math5K
2020-05-14 19:02:31

factors of 30

factors of 30

advanture
2020-05-14 19:02:31

factors of 30

factors of 30

sixoneeight
2020-05-14 19:02:31

factors of 30

factors of 30

MR_67
2020-05-14 19:02:31

they are the factors of 30\

they are the factors of 30\

aidni47
2020-05-14 19:02:31

factors of 30

factors of 30

fasterthanlight
2020-05-14 19:02:31

they are factors of 30

they are factors of 30

TheEpicCarrot7
2020-05-14 19:02:31

because they are the prime factors of 30

because they are the prime factors of 30

IAM_TSSXII
2020-05-14 19:02:31

prime factorization of 30

prime factorization of 30

fasterthanlight
2020-05-14 19:02:31

prime factors of 30

prime factors of 30

Gentoo
2020-05-14 19:02:31

Because those are the prime factors of 30

Because those are the prime factors of 30

Unicorn78
2020-05-14 19:02:31

All prime factors of 30

All prime factors of 30

SharonW
2020-05-14 19:02:31

they are prime factors of 30

they are prime factors of 30

ChrisalonaLiverspur
2020-05-14 19:02:45

because they are the prime factors of 30.

because they are the prime factors of 30.

rrusczyk
2020-05-14 19:02:49

Ah, yes, they're the prime factors of $30.$

Ah, yes, they're the prime factors of $30.$

rrusczyk
2020-05-14 19:02:50

OK, so now we know that $P$ is even, $1$ more than a multiple of $3,$ and $4$ more than a multiple of $5.$ We can write this as

\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$

OK, so now we know that $P$ is even, $1$ more than a multiple of $3,$ and $4$ more than a multiple of $5.$ We can write this as

\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$

rrusczyk
2020-05-14 19:02:58

But... Does this go the other way? That is, must any $P$ that satisfies all three of these give a remainder of $4$ when divided by $30?$

But... Does this go the other way? That is, must any $P$ that satisfies all three of these give a remainder of $4$ when divided by $30?$

edjar
2020-05-14 19:03:12

Yes

Yes

kred9
2020-05-14 19:03:12

yes

yes

rrusczyk
2020-05-14 19:03:17

Let's see! What do the first two give us? ($P=2a$ and $P=3b+1.$)

Let's see! What do the first two give us? ($P=2a$ and $P=3b+1.$)

TThB0501
2020-05-14 19:03:35

b is odd

b is odd

FearlessTaurus
2020-05-14 19:03:35

b is odd

b is odd

MR_67
2020-05-14 19:03:35

b is odd

b is odd

Jomo
2020-05-14 19:03:35

b must be odd.

b must be odd.

fasterthanlight
2020-05-14 19:03:43

b is odd

b is odd

arnav21
2020-05-14 19:03:43

b is odd

b is odd

bacond
2020-05-14 19:03:44

b is odd

b is odd

akpi
2020-05-14 19:03:44

b is odd

b is odd

rrusczyk
2020-05-14 19:03:50

Since $P$ must be even, we know that $b$ must be odd. So, we have $b=2d+1$ for some integer $d,$ and

\[P=3b+1=3(2d+1)+1=6d+4.\]

Since $P$ must be even, we know that $b$ must be odd. So, we have $b=2d+1$ for some integer $d,$ and

\[P=3b+1=3(2d+1)+1=6d+4.\]

rrusczyk
2020-05-14 19:04:23

Notice that "$b$ is odd means $b=2d+1$ for some integer $d$" step -- these sorts of observations can be really helpful in number theory problems!

Notice that "$b$ is odd means $b=2d+1$ for some integer $d$" step -- these sorts of observations can be really helpful in number theory problems!

rrusczyk
2020-05-14 19:04:29

And what happens when we combine this with $P = 5c+4?$

And what happens when we combine this with $P = 5c+4?$

rrusczyk
2020-05-14 19:04:49

We have $P = 6d+4$ and $P=5c+4,$ and $d$ and $c$ are integers...

We have $P = 6d+4$ and $P=5c+4,$ and $d$ and $c$ are integers...

SharonW
2020-05-14 19:04:59

5c=6d

5c=6d

sarahAops2020
2020-05-14 19:04:59

6d=5c

6d=5c

edjar
2020-05-14 19:04:59

5c = 6 d

5c = 6 d

FSMS-BL
2020-05-14 19:04:59

6d = 5c

6d = 5c

E1CAarush
2020-05-14 19:04:59

6d = 5c

6d = 5c

HarleyMathCounts
2020-05-14 19:04:59

5c=6d

5c=6d

peace09
2020-05-14 19:05:13

6d=5c

6d=5c

rrusczyk
2020-05-14 19:05:14

Combining $P=6d+4$ and $P=5c+4$ gives $6d+4=5c+4,$ so $6d=5c.$

Combining $P=6d+4$ and $P=5c+4$ gives $6d+4=5c+4,$ so $6d=5c.$

rrusczyk
2020-05-14 19:05:27

Again, $c$ and $d$ are integers, so what do we know now?

Again, $c$ and $d$ are integers, so what do we know now?

advanture
2020-05-14 19:05:44

c is a multiple of 6

c is a multiple of 6

Math5K
2020-05-14 19:05:44

and d is multiple of 5

and d is multiple of 5

advanture
2020-05-14 19:05:44

d is a multiple of 5

d is a multiple of 5

MR_67
2020-05-14 19:05:44

c is a mult of 6

c is a mult of 6

sixoneeight
2020-05-14 19:05:44

d is a multipe of 5

d is a multipe of 5

nathanqiu
2020-05-14 19:05:44

d is divisible by 5

d is divisible by 5

wlm2
2020-05-14 19:05:44

d is a multiple of 5 and c is a multiple of 6

d is a multiple of 5 and c is a multiple of 6

jupiter314
2020-05-14 19:05:46

d is a multiple of 5, c is a multiple of 6

d is a multiple of 5, c is a multiple of 6

rrusczyk
2020-05-14 19:06:05

$6d=5c$ tells us that $d$ is a multiple of $5,$ and $c$ is a multiple of $6.$

$6d=5c$ tells us that $d$ is a multiple of $5,$ and $c$ is a multiple of $6.$

rrusczyk
2020-05-14 19:06:40

Since $c$ is a multiple of $6,$ we have $c=6f$ for some integer $f.$ What does that give us?

Since $c$ is a multiple of $6,$ we have $c=6f$ for some integer $f.$ What does that give us?

TheEpicCarrot7
2020-05-14 19:07:10

P=30f+4

P=30f+4

FSMS-BL
2020-05-14 19:07:10

30f+4

30f+4

Alculator11
2020-05-14 19:07:10

P=30f+4

P=30f+4

bobjoebilly
2020-05-14 19:07:22

P = 30f+4

P = 30f+4

rrusczyk
2020-05-14 19:07:23

Letting $c = 6f,$ we now have

\[P=5c+4 = 5(6f)+4=30f+4.\]

Letting $c = 6f,$ we now have

\[P=5c+4 = 5(6f)+4=30f+4.\]

nathanqiu
2020-05-14 19:07:33

we already know that

we already know that

Jomo
2020-05-14 19:07:33

which is what we already knew...

which is what we already knew...

peace09
2020-05-14 19:07:33

oof circular reasoning

oof circular reasoning

rrusczyk
2020-05-14 19:07:57

Not quite! When we started, we said that $P=30k+4$ means that the following is true:

Not quite! When we started, we said that $P=30k+4$ means that the following is true:

rrusczyk
2020-05-14 19:08:08

\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$

\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$

rrusczyk
2020-05-14 19:08:15

We JUST WENT THE OTHER WAY!

We JUST WENT THE OTHER WAY!

sixoneeight
2020-05-14 19:08:30

we had to prove that it works the other way

we had to prove that it works the other way

TheEpicCarrot7
2020-05-14 19:08:30

We proved the convers

We proved the convers

rrusczyk
2020-05-14 19:08:45

We started with those three equations and showed that the result is that $P=30f+4$ for some integer $f.$

We started with those three equations and showed that the result is that $P=30f+4$ for some integer $f.$

arnav21
2020-05-14 19:08:58

Wow

Wow

Alculator11
2020-05-14 19:08:58

So we've shown that it's necessary and sufficient

So we've shown that it's necessary and sufficient

MathWiz20
2020-05-14 19:08:58

We proved it

We proved it

akpi
2020-05-14 19:08:58

we proved that it works the other way

we proved that it works the other way

peace09
2020-05-14 19:08:58

we proved the converse... yea

we proved the converse... yea

PShucks
2020-05-14 19:08:59

but how is this helpful?

but how is this helpful?

rrusczyk
2020-05-14 19:09:13

Great question. Why do we like these three new equations?

Great question. Why do we like these three new equations?

rrusczyk
2020-05-14 19:09:23

(What we just worked through here is an example of the Chinese Remainder Theorem, which is really powerful. You might want to check out the Chinese Remainder Theorem later.)

(What we just worked through here is an example of the Chinese Remainder Theorem, which is really powerful. You might want to check out the Chinese Remainder Theorem later.)

cj13609517288
2020-05-14 19:09:37

Since they are easier to use

Since they are easier to use

rocketsri
2020-05-14 19:09:37

They are simpler

They are simpler

rrusczyk
2020-05-14 19:09:47

Exactly, we've broken our problem down to three smaller problems.

Exactly, we've broken our problem down to three smaller problems.

rrusczyk
2020-05-14 19:09:59

Let's start with the first: what is the probability that our product ends up even?

Let's start with the first: what is the probability that our product ends up even?

Trollyjones
2020-05-14 19:10:30

1/2

1/2

sdattilo2002
2020-05-14 19:10:30

1/2

1/2

Jomo
2020-05-14 19:10:30

1/2

1/2

drakewilson
2020-05-14 19:10:30

1/2

1/2

Lux1
2020-05-14 19:10:30

1/2

1/2

tigerjade003
2020-05-14 19:10:30

50%

50%

manokid
2020-05-14 19:10:30

1/2

1/2

mmjguitar
2020-05-14 19:10:30

1/2

1/2

awesomebooks
2020-05-14 19:10:30

1/2

1/2

MathJams
2020-05-14 19:10:30

1/2

1/2

palindrome868
2020-05-14 19:10:30

1/2

1/2

felicitas
2020-05-14 19:10:30

1/2

1/2

rrusczyk
2020-05-14 19:10:51

That's the probability a single number we choose is even.

That's the probability a single number we choose is even.

rrusczyk
2020-05-14 19:11:01

We need the product of all six integers to be even.

We need the product of all six integers to be even.

rrusczyk
2020-05-14 19:11:15

How do we figure that out?

How do we figure that out?

yayy
2020-05-14 19:11:36

1 - 1/2^6

1 - 1/2^6

maxben
2020-05-14 19:11:36

1-(1/2)^6

1-(1/2)^6

NinjaMango
2020-05-14 19:11:36

1 - (1/2)^6 = 63/64

1 - (1/2)^6 = 63/64

Jerry_Guo
2020-05-14 19:11:36

63/64

63/64

Streaks123
2020-05-14 19:11:36

63/64

63/64

MathAcademyIsFun
2020-05-14 19:11:36

1-1/64=63/64

1-1/64=63/64

yangi26
2020-05-14 19:11:39

probability they are ALL odd

probability they are ALL odd

skyguy88
2020-05-14 19:11:40

Complementary counting (odds of it being odd) gives 1 - 1/64 = 63/64

Complementary counting (odds of it being odd) gives 1 - 1/64 = 63/64

logz
2020-05-14 19:11:43

$1-\frac{1}{2^6}$

$1-\frac{1}{2^6}$

melonlord
2020-05-14 19:11:47

find the probability of the product being odd

find the probability of the product being odd

rrusczyk
2020-05-14 19:11:50

In order to be even, we can't have all $6$ chosen numbers be odd.

In order to be even, we can't have all $6$ chosen numbers be odd.

rrusczyk
2020-05-14 19:11:52

All $6$ are odd with probability $\left(\frac12\right)^6=\frac{1}{64},$ so the probability the product is even is

\[1-\frac{1}{64} = \frac{63}{64}.\]

All $6$ are odd with probability $\left(\frac12\right)^6=\frac{1}{64},$ so the probability the product is even is

\[1-\frac{1}{64} = \frac{63}{64}.\]

rrusczyk
2020-05-14 19:12:20

Next in the three things we know about our product -- it must have a remainder of $1$ when divided by $3.$

Next in the three things we know about our product -- it must have a remainder of $1$ when divided by $3.$

rrusczyk
2020-05-14 19:12:54

What must be true about the numbers we choose?

What must be true about the numbers we choose?

Alculator11
2020-05-14 19:12:58

So it can't be divisible by 3

So it can't be divisible by 3

MAthen07
2020-05-14 19:13:03

There can be no number chosen that is divisible by 3.

There can be no number chosen that is divisible by 3.

TheEpicCarrot7
2020-05-14 19:13:15

None of them can be multiples of 3

None of them can be multiples of 3

melonlord
2020-05-14 19:13:15

none of them are divisible by 3

none of them are divisible by 3

aidni47
2020-05-14 19:13:15

none of them can be a mult of 3

none of them can be a mult of 3

rrusczyk
2020-05-14 19:13:20

Ah, that's a good start!

Ah, that's a good start!

rrusczyk
2020-05-14 19:13:33

First up, none of the chosen numbers can be a multiple of $3.$ What's the probability that this happens?

First up, none of the chosen numbers can be a multiple of $3.$ What's the probability that this happens?

fasterthanlight
2020-05-14 19:14:19

64/729

64/729

melonlord
2020-05-14 19:14:19

(2/3)^6

(2/3)^6

mmjguitar
2020-05-14 19:14:19

(2/3)^6

(2/3)^6

ElNoraa
2020-05-14 19:14:19

64/729

64/729

Jomo
2020-05-14 19:14:19

$(2/3)^6 = 64/729$

$(2/3)^6 = 64/729$

rrusczyk
2020-05-14 19:15:41

Each number has a $\frac23$ chance of not being a multiple of $3,$ so the probability that the product is NOT a multiple of $3$ is

\[\left(\frac23\right)^6 = \frac{64}{729}.\]

Each number has a $\frac23$ chance of not being a multiple of $3,$ so the probability that the product is NOT a multiple of $3$ is

\[\left(\frac23\right)^6 = \frac{64}{729}.\]

rrusczyk
2020-05-14 19:16:08

Um, but then our product might be either $1$ or $2$ more than a multiple of $3.$ Now what?

Um, but then our product might be either $1$ or $2$ more than a multiple of $3.$ Now what?

johnnyboy1113
2020-05-14 19:16:24

reduce by 50%

reduce by 50%

szheng0312
2020-05-14 19:16:24

divide by 2

divide by 2

E1CAarush
2020-05-14 19:16:24

Divide by 2

Divide by 2

karthic7073
2020-05-14 19:16:25

divide by 2!

divide by 2!

BakedPotato66
2020-05-14 19:16:27

divide by two- two cases are symmetrical

divide by two- two cases are symmetrical

Math4Life2020
2020-05-14 19:16:29

Times 1/2

Times 1/2

cj13609517288
2020-05-14 19:16:36

are you sure.

are you sure.

rrusczyk
2020-05-14 19:16:45

Good question -- why can we just take half?

Good question -- why can we just take half?

palindrome868
2020-05-14 19:17:14

half of them give us 1

half of them give us 1

MR_67
2020-05-14 19:17:14

half is 1 more, half is 2 more than a mult of 3

half is 1 more, half is 2 more than a mult of 3

Eagle21
2020-05-14 19:17:19

half are 1 more half are 2 more

half are 1 more half are 2 more

rrusczyk
2020-05-14 19:17:22

Convince me.

Convince me.

rrusczyk
2020-05-14 19:17:34

I see a lot of you claiming "symmetry".

I see a lot of you claiming "symmetry".

rrusczyk
2020-05-14 19:17:41

Convince me we have symmetry here.

Convince me we have symmetry here.

rrusczyk
2020-05-14 19:18:54

It is true that the number of numbers from 1 to 900 that are 1 more than a multiple of 3 equals the number of numbers that are 2 more than a multiple of 3.

It is true that the number of numbers from 1 to 900 that are 1 more than a multiple of 3 equals the number of numbers that are 2 more than a multiple of 3.

rrusczyk
2020-05-14 19:19:08

But we care here about the *product of 6 such numbers*.

But we care here about the *product of 6 such numbers*.

rrusczyk
2020-05-14 19:19:34

Why is the product of 6 of these numbers (after we take out the multiples of 3) equally likely to be 1 more or 2 more than a multiple of 3?

Why is the product of 6 of these numbers (after we take out the multiples of 3) equally likely to be 1 more or 2 more than a multiple of 3?

gting
2020-05-14 19:19:51

If we start with 2 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 2 mod 3 or 4 mod 3, which is just 1 mod 3.

If we start with 1 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 1 mod 3 or 2 mod 3, which is 1 mod 3.

So out of all the 4 cases at each step, half are 1 mod 3 and half are 2 mod 3.

If we start with 2 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 2 mod 3 or 4 mod 3, which is just 1 mod 3.

If we start with 1 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 1 mod 3 or 2 mod 3, which is 1 mod 3.

So out of all the 4 cases at each step, half are 1 mod 3 and half are 2 mod 3.

NinjaMango
2020-05-14 19:19:51

(3n+1)(3n+1)=9n^2+6n+1, (3n+2)(3n+1) has a 2 term at the end, and there are 2 cases, and (3n+2)(3n+2) has a four term, which is remainder 1. So two cases of each

(3n+1)(3n+1)=9n^2+6n+1, (3n+2)(3n+1) has a 2 term at the end, and there are 2 cases, and (3n+2)(3n+2) has a four term, which is remainder 1. So two cases of each

rrusczyk
2020-05-14 19:20:14

Great -- let's make this a little more new-student friendly.

Great -- let's make this a little more new-student friendly.

rrusczyk
2020-05-14 19:20:28

We can make a little multiplication table to see what happens when we multiply numbers that are $1$ or $2$ more than a multiple of $3$:

$$\begin{array}{c|cc}

\times & 1 & 2 \\ \hline

1 & 1 & 2 \\

2 & 2 & 1

\end{array}$$

We can make a little multiplication table to see what happens when we multiply numbers that are $1$ or $2$ more than a multiple of $3$:

$$\begin{array}{c|cc}

\times & 1 & 2 \\ \hline

1 & 1 & 2 \\

2 & 2 & 1

\end{array}$$

rrusczyk
2020-05-14 19:21:02

This is just a multiplication table in which we look at the remainder of the product when we multiply two numbers.

This is just a multiplication table in which we look at the remainder of the product when we multiply two numbers.

rrusczyk
2020-05-14 19:21:26

If we multiply two numbers that are 2 more than a multiple of 3, we get a product that is 1 more than a multiple of 3. And so on.

If we multiply two numbers that are 2 more than a multiple of 3, we get a product that is 1 more than a multiple of 3. And so on.

rrusczyk
2020-05-14 19:21:36

Some of you may recognize what we're doing here as "modular arithmetic". If that term is new to you, I recommend you check it out!

Some of you may recognize what we're doing here as "modular arithmetic". If that term is new to you, I recommend you check it out!

rrusczyk
2020-05-14 19:21:41

The products in our table are the remainders we get when we divide the products by $3.$ We see that whether we start with a number that is $1$ or $2$ more than a multiple of $3,$ when we multiply by the next randomly-chosen number, the new product is equally likely to be $1$ or $2$ more than a multiple of $3.$

The products in our table are the remainders we get when we divide the products by $3.$ We see that whether we start with a number that is $1$ or $2$ more than a multiple of $3,$ when we multiply by the next randomly-chosen number, the new product is equally likely to be $1$ or $2$ more than a multiple of $3.$

rrusczyk
2020-05-14 19:21:55

So, if the product of all $6$ numbers is not a multiple of $3,$ it is equally likely to be $1$ or $2$ more than a multiple of $3.$

So, if the product of all $6$ numbers is not a multiple of $3,$ it is equally likely to be $1$ or $2$ more than a multiple of $3.$

rrusczyk
2020-05-14 19:22:06

That is, the probability the product is one more than a multiple of $3$ is $\frac12\cdot \frac{64}{729} = \frac{32}{729}.$

That is, the probability the product is one more than a multiple of $3$ is $\frac12\cdot \frac{64}{729} = \frac{32}{729}.$

rrusczyk
2020-05-14 19:22:10

Almost there! We just have to deal with $5.$

Almost there! We just have to deal with $5.$

rrusczyk
2020-05-14 19:22:13

What do we find there?

What do we find there?

Streaks123
2020-05-14 19:22:52

Same as in this one?

Same as in this one?

cj13609517288
2020-05-14 19:22:58

It is 1/4 of the time. Same logic here except with 4 cases instead of 2.

It is 1/4 of the time. Same logic here except with 4 cases instead of 2.

rrusczyk
2020-05-14 19:23:05

It's basically the same deal as with $3!$ (Phew!)

It's basically the same deal as with $3!$ (Phew!)

Alculator11
2020-05-14 19:23:11

It can't be a multiple of 5

It can't be a multiple of 5

rrusczyk
2020-05-14 19:23:27

None of the numbers chosen can be a multiple of $5$. What's the probability that happens?

None of the numbers chosen can be a multiple of $5$. What's the probability that happens?

melonlord
2020-05-14 19:23:51

(4/5)^6

(4/5)^6

The_Better_Samuel
2020-05-14 19:23:51

(4/5)^6

(4/5)^6

sdattilo2002
2020-05-14 19:23:51

(4/5)^6

(4/5)^6

BakedPotato66
2020-05-14 19:23:51

$(4/5)^6$

$(4/5)^6$

MR_67
2020-05-14 19:23:51

(4/5)^6

(4/5)^6

FearlessTaurus
2020-05-14 19:23:57

(4/5)^6

(4/5)^6

maxben
2020-05-14 19:23:59

(4/5)^6

(4/5)^6

jai123
2020-05-14 19:23:59

(4/5)^6

(4/5)^6

rrusczyk
2020-05-14 19:24:01

None of the chosen numbers can be multiples of $5,$ which happens with probability $(4/5)^6.$

None of the chosen numbers can be multiples of $5,$ which happens with probability $(4/5)^6.$

rrusczyk
2020-05-14 19:24:06

Then, we can make a multiplication table to look at possible remainders of products:

$$\begin{array}{c|cccc}

\times & 1&2&3&4 \\ \hline

1 & 1&2&3&4 \\

2 & 2&4&1&3 \\

3 & 3&1&4&2 \\

4 & 4&3&2&1 \end{array}$$

Then, we can make a multiplication table to look at possible remainders of products:

$$\begin{array}{c|cccc}

\times & 1&2&3&4 \\ \hline

1 & 1&2&3&4 \\

2 & 2&4&1&3 \\

3 & 3&1&4&2 \\

4 & 4&3&2&1 \end{array}$$

rrusczyk
2020-05-14 19:24:18

(You can work through that multiplication table on your own later ;) )

(You can work through that multiplication table on your own later ;) )

Jomo
2020-05-14 19:24:26

huh they are all equal...

huh they are all equal...

Kruxe
2020-05-14 19:24:26

that has a cool pattern lol

that has a cool pattern lol

MathJams
2020-05-14 19:24:29

all happen equally amounts of times

all happen equally amounts of times

rrusczyk
2020-05-14 19:24:33

Exactly!

Exactly!

rrusczyk
2020-05-14 19:24:44

So...

So...

MR_67
2020-05-14 19:24:47

1/4 that it is 4 mod 5

1/4 that it is 4 mod 5

melonlord
2020-05-14 19:24:47

it's 1/4

it's 1/4

MR_67
2020-05-14 19:24:56

1/4

1/4

aidni47
2020-05-14 19:24:56

1/4

1/4

sarahAops2020
2020-05-14 19:24:56

*1/4

*1/4

rrusczyk
2020-05-14 19:25:14

Yep. $1/4$ of these products will end up being 4 more than a multiple of 5.

Yep. $1/4$ of these products will end up being 4 more than a multiple of 5.

rrusczyk
2020-05-14 19:25:16

So, the probability our product is $4$ more than a multiple of $5$ is

\[\frac14\cdot \frac{4^6}{5^6} = \frac{4^5}{5^6}.\]

So, the probability our product is $4$ more than a multiple of $5$ is

\[\frac14\cdot \frac{4^6}{5^6} = \frac{4^5}{5^6}.\]

cj13609517288
2020-05-14 19:25:23

thank goodness we're done. My brain is basically ded.

thank goodness we're done. My brain is basically ded.

rrusczyk
2020-05-14 19:25:26

Mine too.

Mine too.

BakedPotato66
2020-05-14 19:25:32

yay symmetry is fun!!!

yay symmetry is fun!!!

rrusczyk
2020-05-14 19:25:35

Yes!

Yes!

rrusczyk
2020-05-14 19:25:39

Now, how do we finish the problem?

Now, how do we finish the problem?

sdattilo2002
2020-05-14 19:26:05

now we just multiply

now we just multiply

Rsar12
2020-05-14 19:26:05

then multiply all the probabilities together

then multiply all the probabilities together

sarahAops2020
2020-05-14 19:26:05

multipy

multipy

ElNoraa
2020-05-14 19:26:05

multiply all together

multiply all together

MathJams
2020-05-14 19:26:05

multiply them?

multiply them?

fasterthanlight
2020-05-14 19:26:05

multiply the probabilities?

multiply the probabilities?

coldcrazylogic
2020-05-14 19:26:05

multiply everything

multiply everything

sixoneeight
2020-05-14 19:26:08

multiply them!

multiply them!

MAthen07
2020-05-14 19:26:08

Multiply em' all together.

Multiply em' all together.

rrusczyk
2020-05-14 19:26:12

Our three events are independent (the remainder we get upon dividing by $2,3,$ or $5$ does not depend on the other two remainders -- think later about why). So, we multiply our three probabilities.

Our three events are independent (the remainder we get upon dividing by $2,3,$ or $5$ does not depend on the other two remainders -- think later about why). So, we multiply our three probabilities.

rrusczyk
2020-05-14 19:26:14

You might want to simplify this a little bit before entering it into your calculator: $$\frac{63}{64} \cdot \frac{32}{729} \cdot \dfrac{4^5}{5^6} = \dfrac{7 \cdot 512}{81 \cdot 5^6}.$$

You might want to simplify this a little bit before entering it into your calculator: $$\frac{63}{64} \cdot \frac{32}{729} \cdot \dfrac{4^5}{5^6} = \dfrac{7 \cdot 512}{81 \cdot 5^6}.$$

rrusczyk
2020-05-14 19:26:21

What do we get?

What do we get?

jai123
2020-05-14 19:26:36

calculator time?

calculator time?

rrusczyk
2020-05-14 19:26:38

For sure!

For sure!

Jerry_Guo
2020-05-14 19:26:56

0.283%

0.283%

MR_67
2020-05-14 19:26:56

0.283%

0.283%

Kruxe
2020-05-14 19:27:11

0.283% oops

0.283% oops

BakedPotato66
2020-05-14 19:27:11

0.283 %

0.283 %

prajna1225
2020-05-14 19:27:11

0.283%

0.283%

onedance
2020-05-14 19:27:11

0.283 percent

0.283 percent

acornfirst
2020-05-14 19:27:11

0.283%

0.283%

bronzetruck2016
2020-05-14 19:27:11

0.283%

0.283%

rrusczyk
2020-05-14 19:27:15

This computes to be $0.002831\ldots$.

This computes to be $0.002831\ldots$.

rrusczyk
2020-05-14 19:27:18

As a percent to the nearest thousandth, it is $\boxed{0.283}\%$.

As a percent to the nearest thousandth, it is $\boxed{0.283}\%$.

casp
2020-05-14 19:27:26

"rrusczyk: We're not sure! Maybe 90 minutes or so."

"rrusczyk: We're not sure! Maybe 90 minutes or so."

rrusczyk
2020-05-14 19:27:41

I was right! I think I just did 90 minutes. DPatrick did the rest

I was right! I think I just did 90 minutes. DPatrick did the rest

rrusczyk
2020-05-14 19:27:58

That's it for the problems tonight.

That's it for the problems tonight.

rrusczyk
2020-05-14 19:28:14

Be sure to keep playing around with the others -- there are still a bunch more to play with!

Be sure to keep playing around with the others -- there are still a bunch more to play with!

Quaoar
2020-05-14 19:28:18

This was so much fun and so enlightening! Thank you amysz, bedwards, Binomial-theorem, corinne, dcouchman, DPatrick, kdej22, LadyBlue 32, nbasrl, rponda, rrusczyk, sarahtrebatleder, tuvie, and vapodaca!

This was so much fun and so enlightening! Thank you amysz, bedwards, Binomial-theorem, corinne, dcouchman, DPatrick, kdej22, LadyBlue 32, nbasrl, rponda, rrusczyk, sarahtrebatleder, tuvie, and vapodaca!

edgymemelord
2020-05-14 19:28:23

ITS BEEN 4 HOURS?!?!

ITS BEEN 4 HOURS?!?!

rrusczyk
2020-05-14 19:28:27

No kidding!

No kidding!

donguri
2020-05-14 19:28:31

is this the longest ever math jam?

is this the longest ever math jam?

rrusczyk
2020-05-14 19:28:36

Probably.

Probably.

TMSmathcounts737
2020-05-14 19:28:39

this lasted for 4 hours and 13 minutes.

this lasted for 4 hours and 13 minutes.

rrusczyk
2020-05-14 19:28:52

Should have been one minute less than one hour less.

Should have been one minute less than one hour less.

lilcritters
2020-05-14 19:28:59

Will there be a similar Math Jam next year?

Will there be a similar Math Jam next year?

rrusczyk
2020-05-14 19:29:07

Maybe, but hopefully not for the same reason!

Maybe, but hopefully not for the same reason!

razhder
2020-05-14 19:29:10

How many people do you think it would take to break the schoolhouse?

How many people do you think it would take to break the schoolhouse?

rrusczyk
2020-05-14 19:29:17

More than 1400, we learned today!

More than 1400, we learned today!

rrusczyk
2020-05-14 19:29:20

As a reminder, there are two more MATHCOUNTS Week events tomorrow:

As a reminder, there are two more MATHCOUNTS Week events tomorrow:

rrusczyk
2020-05-14 19:29:23

**Q&A with John Urschel**at 3:14pm ET/12:14pm PT on the MATHCOUNTS YouTube channel.
rrusczyk
2020-05-14 19:29:25

**The World's Largest Countdown Round**at 6:14pm ET/3:14pm PT right back in this room. See you there!
winner320
2020-05-14 19:29:31

goodbye everyone

goodbye everyone

MokshaParam
2020-05-14 19:29:31

yay!

yay!

mmjguitar
2020-05-14 19:29:31

Thank you!!

Thank you!!

aidni47
2020-05-14 19:29:31

Thanks so much for doing this!

Thanks so much for doing this!

Math5K
2020-05-14 19:29:50

Thank you so much!

Thank you so much!

mfro24
2020-05-14 19:29:50

Thank you so much all!

Thank you so much all!

mathicorn
2020-05-14 19:29:50

I really liked this

I really liked this

akpi
2020-05-14 19:29:50

thank you rrusczyk

thank you rrusczyk

mfro24
2020-05-14 19:29:50

I can't believe it's been this long!

I can't believe it's been this long!

onedance
2020-05-14 19:29:52

will you post the transcript?

will you post the transcript?

DPatrick
2020-05-14 19:29:52

Thanks for coming!

Thanks for coming!

rrusczyk
2020-05-14 19:29:57

We will indeed!

We will indeed!

rrusczyk
2020-05-14 19:30:04

Thank you everybody!

Thank you everybody!

EpicGirl
2020-05-14 19:30:19

Thank You!

Thank You!

dolphin7
2020-05-14 19:30:19

Thank You!

Thank You!

mohanty
2020-05-14 19:30:19

Thank you so much!

Thank you so much!

SharonW
2020-05-14 19:30:19

THANK YOU!

THANK YOU!

Apple321
2020-05-14 19:30:19

thanks

thanks

arirah9
2020-05-14 19:30:19

Thank you so much!!!

Thank you so much!!!

huela
2020-05-14 19:30:19

Thank you!!

Thank you!!

karthic7073
2020-05-14 19:30:25

Thanks for teaching!

Thanks for teaching!

Coolpeep
2020-05-14 19:30:25

bye! thank you!

bye! thank you!

MathWiz20
2020-05-14 19:30:25

Bye! Yay! Thank you so much!

Bye! Yay! Thank you so much!

johnnyboy1113
2020-05-14 19:30:25

Yay! Thanks a lot

Yay! Thanks a lot

Beastboss
2020-05-14 19:30:28

Thank you!

Thank you!

PracticingMath
2020-05-14 19:30:40

Thank you!

Thank you!

Shubhashubha
2020-05-14 19:30:43

Thank YOU

Thank YOU

brainyharry
2020-05-14 19:30:45

where will you post the transcript

where will you post the transcript

rrusczyk
2020-05-14 19:30:59

https://artofproblemsolving.com/school/mathjams-transcripts

https://artofproblemsolving.com/school/mathjams-transcripts

ohjoshuaoh
2020-05-14 19:31:30

Thank you everyone!!!

Thank you everyone!!!

akpi2
2020-05-14 19:31:30

TTHHAANNKKSS!!

TTHHAANNKKSS!!

advanture
2020-05-14 19:31:30

THANKS SO MUCH

THANKS SO MUCH

rrusczyk
2020-05-14 19:32:06

Thanks also to our superstar assistants!

Thanks also to our superstar assistants!

Juneybug
2020-05-14 19:32:25

how do you pronounce your last name rusczyk?

how do you pronounce your last name rusczyk?

rponda
2020-05-14 19:32:27

Thank you.

Thank you.

rrusczyk
2020-05-14 19:32:33

RUH-sick (just like it looks )

RUH-sick (just like it looks )

TheEpicCarrot7
2020-05-14 19:32:58

Lol imagine we ended at 6:28

Lol imagine we ended at 6:28

mohanty
2020-05-14 19:32:58

WE STARTED AT PI TIME(EST) AND ENDED AT TAU TIME(CST)

WE STARTED AT PI TIME(EST) AND ENDED AT TAU TIME(CST)